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Enthalpy - class-XII

Description: enthalpy
Number of Questions: 37
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Tags: thermodynamics enthalpy changes chemical thermodynamics chemistry
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Heat of reaction at constant pressure is called

  1. Enthalpy.

  2. Entropy

  3. Free energy

  4. None of these


Correct Option: A
Explanation:

Enthalpy : Chemical reactions are generally carried out at constant pressure (atmospheric pressure) and heat of reaction at constant pressure is called Enthalpy (H) as:
$H = U + PV$ (By definition).

The relationship between enthalpy and internal energy change is

  1. $\,\Delta U = \Delta H + P\Delta V$

  2. $\,\Delta H = \Delta U + P\Delta V$

  3. $\,\Delta H = \Delta U - P\Delta V$

  4. $\,P\Delta V = \Delta U + \Delta H$


Correct Option: B
Explanation:
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as :
$ H = U + PV$ ( By definition )
$ \Delta H = \Delta U + \Delta (PV)$
$ \Delta H = \Delta  U + P\Delta V $ (at constant pressure) combining with first law.
$ \Delta H = q _{p}$

An athelete is given 100 g of glucose $(C _6H _{12}O _6)$ of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, Determine the weight of water he would need to perspire. (The enthalpy of evaporation of water is 44 kJ/mole.)

  1. 319 gm

  2. 323 gm

  3. 342 gm

  4. 312


Correct Option: A
Explanation:

 Net amount of energy given to athlete $= 1560 kJ$

$1560 × 50/100$ Energy lost in an event $= 780 kJ$
 Energy left out $= 1560 – 780 kJ = 780 kJ$
 Now, consider the evaporation of water $H _2O(l) \rightarrow H _2O(g)$; $\Delta H = 44 kJ mole^{–1}$
 Thus, for consumption of $44 kJ$ of energy the amount of water evaporated $=1mole= 18 g$ 
For consumption of $780 kJ$ of energy the amount of water to be evaporated $ \frac {18 \times  780}{44} = 319·09 g$

Enthalpy of the system is given as :

  1. $\displaystyle H+PV$

  2. $\displaystyle U+PV$

  3. $\displaystyle U-PV$

  4. $\displaystyle H-PV$


Correct Option: B
Explanation:

Enthalpy: Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy $(H)$ as :
$H = U + PV$ 

Under which of the following condition is the relation $\Delta H = \Delta U + P\Delta V$ valid for a closed system at

  1. constant pressure

  2. constant temperature

  3. constant temperature and pressure

  4. constant temperature, pressure and composition


Correct Option: A
Explanation:
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been 
found useful to define a new state function Enthalpy(H) as:
$ H = U + PV$ (By definition)
$ \Delta H = \Delta U + \Delta (PV)$
$ \Delta H = \Delta U + P\Delta V$  (at constant pressure) combining with first law. 
$ \Delta H = q _p $

In the hydrogenation of ethylene,$\Delta n$ is equal to:

  1. 1

  2. -2

  3. -1

  4. 2


Correct Option: C
Explanation:

$C _{2}H _{4}(g)+H _{2}(g)\rightarrow C _{2}H _{6}(g)$

$\Delta n=1-2$

$=$ - 1


Option C is correct.

For the reaction,

$C(s)+O {2}(g)\rightarrow CO _{2}(g);$

$\Delta n$ value is _________.

  1. Zero

  2. +1

  3. -1

  4. unpredictable


Correct Option: A
Explanation:

$C(s)+O _{2}(g)\rightarrow CO _{2}(g)$


$\Delta n _{g}=1-1$

$=0$ 

$=$ difference of moles of gaseous substances

Option A is correct.

the $\triangle { H } _{ 1 }\quad $ and $\triangle { H } _{ 2 }\quad $ $(in kJ mol^-1)$ and the $\triangle {egH } $  $(in kJ mol^-1)$ of a few 3 elements are given below:
Which of the element is likel;y to be:

Elements $\triangle { H } _{ 1 }\quad $ $\triangle { H } _{ 2 }\quad $  $\triangle {egH } $
I 520 7300 -60
II 419 3051 -48
III 1681 3374 -328
IV 1008 1846 -295
V 2372 5251 +48
VI 738 1451 -40
  1. the least reactive element

  2. the most reactive mental

  3. the most reactive non- mental

  4. the least reactive non-mental


Correct Option: A

What is abbreviated as '$H$'?

  1. Standard voltaic potential

  2. Entropy

  3. Enthalpy

  4. Reaction rate


Correct Option: C
Explanation:

$H$ is enthalpy.

Enthalpy is equivalent to the total heat content of a system.
$\Delta H=\Delta U=\Delta (PV)$

Which of the following statements are correct?

  1. absolute value of enthalpy cannot be determined.

  2. absolute value of internal energy cannot be determined.

  3. absolute value of entropy can be determined.

  4. internal energy, enthalpy, and entropy are intensive properties.


Correct Option: A,B
Explanation:

Properties which depend on the amount of the substance (or substances)

present in the system are called extensive propterties. e.g. Mass,

volume, heat capacity, internal energy, entropy, Gibb's free energy (G),

surface area etc. These properties will change with change in the

amount of matter present in the system.
The absolute value of

internal energy cannot be determined because it is not possible to

determine the exact values for the constituent energies such as

translational, vibrational, rotational energies, etc. However, we can

determine the change in internal energy (U) of the system when it

undergoes a change from initial state (Ul) to final state (Uf).
Enthalpy

is not a matter. It doesn't have mass and it doesn't occupy space.

Therefore, we cannot measure its absolute value (like energy).
We can only measure the changes because the energy of the universe is constant.

Enthalpy of the system is given as

  1. $\,H + PV$

  2. $\,U + PV$

  3. $\,U - PV$

  4. $\,H - PV$


Correct Option: B
Explanation:
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as :
$H=U+PV $ (By definition)
$\Delta H=\Delta U+\Delta (PV)$
$\Delta H=\Delta U+P\Delta V$ (at constant pressure) combining with first law.
$\Delta H=q _{p}$

Which of the following reactions have same heat of reaction at constant $P$ and constant volume as well?

  1. $2NO(g)\longrightarrow N _2(g)+O _2(g)$

  2. $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$

  3. $Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$

  4. $H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$


Correct Option: A,C,D
Explanation:

The following reactions have the same heat of reaction at constant $P$ and constant volume as well.


$2NO(g)\longrightarrow N _2(g)+O _2(g)$
$Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
$H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products is the same.

$\because \Delta n _g = 0$

However, for the reaction $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$, heat of reaction at constant $P$ and constant volume are different.

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products are different.

For which reaction will $\Delta H = \Delta U$?


 Assume each reaction is carried out in an open container.

  1. $H _2(g) + Br _2(g)\longrightarrow 2HBr(g)$

  2. $C(s) + 2H _2O(g)\longrightarrow 2H _2(g) + CO _2(g)$

  3. $4CO(g) + 2O _2(g)\longrightarrow 4CO _2(g)$

  4. $2PCl _5(g)\longrightarrow 2PCl _3(g) + 2Cl _2(g)$


Correct Option: A
Explanation:

$\Delta H = \Delta U + \Delta nRT$, where $\Delta n =$ Change in number of moles.


$H _2 + Br _2 \to 2HBr$


 $\Delta n = 0$.  

$\therefore$ $\Delta H = \Delta U$

Hence, option A is correct

In which of the following reactions, $\Delta H > \Delta U$?

  1. $H _2(g) + I _2(g)\rightarrow 2HI(g)$

  2. $PCl _5(g)\rightarrow PCl _3(g) + Cl _2(g)$

  3. $2H _2O _2(l)\rightarrow 2H _2O(l) + O _2(g)$

  4. $C(s) + O _2(g)\rightarrow CO _2(g)$


Correct Option: B,C
Explanation:

As we know,
$\Delta H = \Delta U + \Delta nRT$, where $\Delta n = n _P - n _R$ (n = number of moles)
$H _2(g) + I _2(g)\rightarrow 2HI(g)$      $\Delta n = 0$
$PCl _5(g)\rightarrow PCl _3(g) + Cl _2(g)$                     $\Delta n = 1$
$2H _2O _2(l)\rightarrow 2H _2O(l) + O _2(g)$                 $\Delta n = 1$
$C(s) + O _2(g)\rightarrow CO _2(g)$                  $\Delta n = 0$

Enthalpy of the system is given as:

  1. $U + PV$

  2. $H = PV$

  3. $U - PV$

  4. $ H - V$


Correct Option: A
Explanation:

Enthalpy is a system is defined as,

$H=U+PV$
Hence, option A is correct

Match List I with List II and select the answer from the given codes. 


List I                                                                               List II
A. $C(s) + O _2(g)\longrightarrow CO _2(g)$                       1. $\Delta H = \Delta U + RT$
B. $N _2(g) + 3H _2(g)\longrightarrow 2NH _3(g)$                2. $\Delta H = \Delta U$
C. $NH _4HS(s)\longrightarrow NH _3(g) + H _2S(g)$       3. $\Delta H =\Delta U - 2RT$
D. $PCl _5(g)\longrightarrow PCl _3(g) + Cl _2(g)$              4. $\Delta H = \Delta U + 2RT$
E. $2SO _2(g) + O _2(g)\longrightarrow 2SO _3(g)$              5. $\Delta H = \Delta U - RT$

  1. $A-1, B-2, C-3, D-4, E-5$

  2. $A-5, B-2, C-3, D-4, E-1$

  3. $A-1, B-3, C-4, D-2, D-5$

  4. $A-2, B-3, C-4, D-1, E-5$


Correct Option: D
Explanation:

As we know,


$\Delta H = \Delta U + \Delta nRT$
$C(s) + O _2(g)\longrightarrow CO _2(g)$       
$\Delta n =0$ so         $\Delta H = \Delta U$

$N _2(g) + 3H _2(g)\longrightarrow 2NH _3(g)$    
$\Delta n =0$ so  $\Delta H =\Delta U - 2RT$


$NH _4HS(s)\longrightarrow NH _3(g) + H _2S(g)$     
$\Delta n =+2$ so  $\Delta H =\Delta U + 2RT$

$PCl _5(g)\longrightarrow PCl _3(g) + Cl _2(g)$         
$\Delta n = 1$ so  $\Delta H =\Delta U + RT$

$2SO _2(g) + O _2(g)\longrightarrow 2SO _3(g)$
$\Delta n = -1 $ so  $\Delta H =\Delta U - RT$     

$H _2(g) + I _2(g)\longrightarrow 2HI(g)$ 


For this reaction, relate $\Delta H$ and $\Delta U$.

  1. $\Delta H$ =$\Delta U$

  2. $\Delta H$ > $\Delta U$

  3. $\Delta H$ < $\Delta U$

  4. None of these


Correct Option: A
Explanation:

As we know,
$\Delta H$ =$\Delta U+ \Delta n _g RT$
here, 
$H _2(g) + I _2(g)\longrightarrow 2HI(g)$ 
$\Delta n _g = 0$
so,
$\Delta H$ =$\Delta U$

Heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction $N _2 + 3H _2 \longrightarrow  2NH _3$ differ $(\Delta H- \Delta U)$ by the amount:

  1. $2RT$

  2. $-2RT$

  3. $3RT$

  4. $RT$


Correct Option: B
Explanation:

The heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction 


$N _2 + 3H _2 \longrightarrow  2NH _3$

$\Delta H = \Delta U + (\Delta n _g)\times RT$

$\Delta H = \Delta U + (-2)\times RT$

This is because the change in the number of moles of gaseous products and the gaseous reactants in the above reaction is -2.

$\Delta H - \Delta U = (-2)\times RT$

Hence, option B is correct.

If $\Delta E$ is the heat of reaction for
${C} _{2}{H} _{5}OH\left(l\right) + 3{O} _{2}\left(g\right) \longrightarrow 2C{O} _{2}\left(g\right) + 3{H} _{2}O\left(l\right)$
at constant volume, the $\Delta H$ (heat of reaction at constant pressure), at constant temperature is:

  1. $\Delta H = \Delta E + RT$

  2. $\Delta H = \Delta E - RT$

  3. $\Delta H = \Delta E - 2RT$

  4. $\Delta H = \Delta E + 2RT$


Correct Option: B
Explanation:

We know that,
$\Delta H = \Delta E + \Delta nRT$
where,
$\Delta n =$ number of moles of gaseous products - number of moles of gaseous reactants
$= 2-3=-1$
So, $\Delta H = \Delta E - RT$

For gaseous reactions, if $\Delta H$ is the change in enthalpy and $\Delta U$ that in internal energy, then

  1. $\Delta H$ is always greater than $\Delta U$

  2. $\Delta H$ is always less than $\Delta U$

  3. $\Delta H > \Delta U$ only if the number of mole of the products is less than that of the reactants.

  4. $\Delta U < \Delta H$ only if the number of mole of the reactants is less than that of the products.


Correct Option: C,D
Explanation:

As we know,
$\,\Delta H = \Delta U + \Delta nRT$
$\Delta n = n _{ P } - n _{ R }$
so
$\Delta U < \Delta H$ only if the number of mole of the reactants is less than that of the products.
$\Delta H > \Delta U$ only if the number of mole of the products is less than that of the reactants.

At constant pressure:

  1. pdv=d(pv)

  2. dQ=du+d(pv)

  3. dQ=d(u+pv)

  4. all of the mentioned


Correct Option: D

For the combustion reaction at $298\,K$
$2Ag(s) +1/2O _2(g)\longrightarrow 2Ag _2O(s)$ Which of the following alternative is correct?

  1. $\Delta H = \Delta U$

  2. $\Delta H > \Delta U$

  3. $\Delta H < \Delta U$

  4. $\Delta H and \Delta U$ bear no relation with each other


Correct Option: C
Explanation:

As we know,
$\,\Delta H = \Delta U + \Delta nRT$
$\Delta n = n _{ P } - n _{ R }$
$\therefore\Delta n = 2 \displaystyle - \frac { 5 }{ 2 } =\displaystyle -\frac{ 1 }{ 2 }$
$\therefore\Delta H = \Delta U \displaystyle - \frac { 1 }{ 2 }RT$
$\therefore\Delta U = \Delta H +\displaystyle \frac{ 1 }{ 2 }RT$
or $\therefore\Delta U > \Delta H$

Hess's law is based on:

  1. Law of conservation of mass

  2. Law of conservation of energy

  3. Second law of thermodynamics

  4. None of the above


Correct Option: B
Explanation:

The Hess's law states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or in several steps. Hess's law is now understood as an expression of the principle of conservation of energy, also expressed in the first law of thermodynamics, and the fact that the enthalpy of a chemical process is independent of the path taken from the initial to the final state (i.e. enthalpy is a state function).

For a particular reaction ; $\Delta E$ $=\Delta H + P\Delta $V.
  1. True

  2. False


Correct Option: B
Explanation:

The given statement is false.
For a particular reaction $\displaystyle \Delta H = \Delta  E + P \Delta  V$
The change in enthalpy is equal to the sum of the change in the internal energy and the pressure volume work. This expression is applicable at constant pressure.

Select the correct option(s):

  1. $q=nC _{v}\mathrm{d} T$ is applicable to all substances during heating/cooling at constant 'v'.

  2. $q=nC _{v}\mathrm{d} T$ is applicable to ideal gas during heating/cooling at constant 'v'.

  3. $\mathrm{d} U=nC _{v}\mathrm{d} T$ is applicable for real gas at constant 'v'

  4. $\mathrm{d} U=nC _{v}\mathrm{d} T$ is applicable for ideal gas at constant 'v' only


Correct Option: A,B
Explanation:

An isochoric process is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant. 


Work done by system is given as $W=P\Delta V$

So workdone by the ideal and real gases will be zero in all cases.

From first law of thermodynamics, $Q=W+\Delta U$
$\Longrightarrow Q=\Delta U$

$\Delta U=n{ C } _{ v }\Delta T$ is applicable for all conditions whether the volume is constant or not.

Enthalpy change of a reaction will be equal to:

  1. $\Delta U + P\Delta V$

  2. $\Delta U + V\Delta P$

  3. $\Delta U + \Delta (PV)$

  4. $\Delta U + (\Delta n _{ g } )\Delta (PV)$


Correct Option: C
Explanation:

The relationship between the enthalpy $H$, internal energy $E$, pressure $P$ and volume $V$ is $H=E+PV$
Hence, the expression for the enthalpy change becomes $\Delta H = \Delta U + \Delta (PV)$
Hence, the enthalpy change of a reaction will be equal to $\Delta U + \Delta (PV)$.

In an isothermal process:

  1. $q=0$ and $\Delta E=0$

  2. $q\neq 0$ and $\Delta E=0$

  3. $q=0$ and $\Delta E\neq 0$

  4. $q\neq 0$ and $\Delta E\neq 0$


Correct Option: B
Explanation:

An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. In other words, in an isothermal process, the value ΔT = 0 and therefore the change in internal energy ΔU = 0 (only for an ideal gas) but Q ≠ 0, while in an adiabatic process, ΔT ≠ 0 but Q = 0.

Two reactions are given below:

(i) $CO _{(g)} + \frac{1}{2}O _2{(g)} \rightarrow CO _2{(g)}$

(ii) $ Ag _2O _{(s)} \rightarrow 2Ag _{(s)} + \frac{1}{2}O _2{(g)}$

Which of the following statements is true?

  1. For (i) $\vartriangle$H < $\vartriangle$E and for (ii) $\vartriangle$H > $\vartriangle$E

  2. For (i) $\vartriangle$H > $\vartriangle$E and for (ii) $\vartriangle$H < $\vartriangle$E

  3. For both (i) and (ii) $\vartriangle$H > $\vartriangle$E

  4. For both (i) and (ii) $\vartriangle$H < $\vartriangle$E


Correct Option: A
Explanation:

$\Delta H=\Delta E+\Delta { n } _{ g }RT$ 


(i) $CO(g) + \cfrac { 1 }{ 2 } { O } _{ 2 }(g) \longrightarrow  C{ O } _{ 2 }(g)$

$ \quad   \Delta { n } _{ g }=1-\cfrac { 1 }{ 2 } -1=-\cfrac { 1 }{ 2 } <0$

$\therefore \Delta H<\Delta E$

(ii) ${ Ag } _{ 2 }O(s) \longrightarrow  2Ag(s) + \cfrac { 1 }{ 2 } { O } _{ 2 }(g)$

$\quad  \Delta { n } _{ g }=\cfrac { 1 }{ 2 } +0-0=\cfrac { 1 }{ 2 } >0$ 

$\therefore \Delta H>\Delta E$

So, the correct option is A.

Which of the following is abbreviated as $H$?

  1. Standard voltaic potential

  2. Entropy

  3. Enthalpy

  4. Reaction rate

  5. Gibbs free energy


Correct Option: C
Explanation:

Enthalpy is abbreviated as $H$.
The abbreviations for other terms are:
Standard voltaic potential: $V$
Entropy: $S$ 
Gibbs free energy: $G$

Total enthalpy of a system H is given by:
  1. $H=h/m$

  2. $H=m/h$

  3. $H=m\times h$

  4. None of the mentioned


Correct Option: C

A mixture of 2 mole of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. If $\Delta H$ is the enthalpy change and $\Delta U$ is the change in internal energy, then:

  1. $\Delta H >\Delta U$

  2. $\Delta H <\Delta U$

  3. $\Delta H =\Delta U$

  4. can't be predicted


Correct Option: B
Explanation:

A mixture of 2 moles of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. 


 $2CO(g)+ O _2(g) \longrightarrow 2CO _2(g)$

 $\Delta n _g =2-3=-1$

If $\Delta H$ is the enthalpy change and $\Delta U$ is the change in internal energy, then:

 $\Delta H=\Delta U+\Delta n _gRT=\Delta U-RT$

Hence,  $\Delta H <\Delta U$

Predict $\Delta H>\Delta U$ or $\Delta H<\Delta U$.

a. $C(graphite) + O _2(g)\longrightarrow CO _2(g)$

b. $PCl _5(g)\longrightarrow PCl _3(g) + Cl _2(g)$

  1. a. $\Delta H = \Delta U$;  b. $\Delta H>\Delta U$

  2. a. $\Delta H > \Delta U$;  b. $\Delta H=\Delta U$

  3. a. $\Delta H >\Delta U$;  b. $\Delta H<\Delta U$

  4. None of these


Correct Option: A
Explanation:
We know,

$\Delta H= \Delta U + \Delta n _gRT$
a. $\Delta H = \Delta U$ because $\Delta n _g = 0$.   

b. $\Delta H>\Delta U$ because $\Delta n _g = 1$ and decomposition of $PCl _5$ is endothermic reaction.

Hence, option A is correct 

Water is boiled under a pressure of 1.0atm. When an electric current of 0.50A from a 12V supply is passed for 300 second through a resistance in thermal contact with it, it is found that 0.789g of water is vapourized. The molar internal energy change at boiling point (373.15K) is

  1. $\displaystyle =37.9kJ{ mol }^{ -1 }$

  2. $\displaystyle =27.5kJ{ mol }^{ -1 }$

  3. $\displaystyle =47.5kJ{ mol }^{ -1 }$

  4. $\displaystyle =17.5kJ{ mol }^{ -1 }$


Correct Option: A
Explanation:

The vaporization occurs at constant pressure therefore the enthalpy change is equal to the work done by the heater:

$\displaystyle \Delta { H }^{ \prime  }=0.50\times 12\times 300\quad (\Delta H=i\times V\times t)$
$\displaystyle =1800J$
$\displaystyle =+1.8kJ$

$\displaystyle \therefore $Molar enthalpy of vaporization, $\displaystyle \Delta H=\frac { \Delta { H }^{ \prime  } }{ mole\ of\ { H } _{ 2 }{ O } } $
$\displaystyle =\frac { 1.8 }{ \left( \frac { 0.789 }{ 18 }  \right)  } $

$\displaystyle =41.06\ kJ{ mol }^{ -1 }$

Also, $\displaystyle \Delta H=\Delta U+P\Delta V$

$\displaystyle =\Delta U+\Delta { n } _{ g }RT$

$\displaystyle =\Delta U+RT$ $\displaystyle ((\because { H } _{ 2 }O\left( l \right) \rightleftharpoons { H } _{ 2 }O\left( g \right) ,\Delta { n } _{ g }=1)$

$\displaystyle \therefore \Delta U=$ molar internal energy change

$\displaystyle\Delta U= \Delta H-RT$

$\displaystyle =41.06-(8.314\times { 10 }^{ -3 }\times 373.15)$

$\displaystyle =37.96\ kJ{ mol }^{ -1 }$

A piston exerting a pressure of 1.0 atmosphere rests on the surface of water at $100^{\circ}C$. The pressure is reduced to smaller extent and as a result 10 g of water evaporates and absorbs 22.2 kJ of heat. The change in internal energy is:

  1. 18.24 kJ

  2. 20.477 kJ

  3. 22.05 kJ

  4. 23.923 kJ


Correct Option: B
Explanation:

The relationship between the enthalpy change and the change in the internal energy is
 $\Delta H=\Delta U+P\Delta V=\Delta U+\Delta n _gRT$
$\Delta U=\Delta H-\Delta n _gRT$
Substitute values in the above expression.
$\Delta U=\displaystyle 22.2-\frac{10}{18}\times 8.314\times 10^{-3}\times 373$
                       $=20.477 :kJ$
The change in internal energy is 20.477 kJ

Enthalpy is_________.

  1. Energy of reaction

  2. Tendency for loss of energy

  3. Tendency for randomness

  4. Use of energy in overcoming entropy


Correct Option: A
Explanation:

Enthalpy is the measurement of energy in a thermodynamic system. It is equal to the internal energy of the system plus the product of pressure and volume. The thermal change that occurs in a chemical reaction is only due to the difference in the sum of internal energy of the products and the sum of the internal energy of reactants. At constant pressure, the heat of the reaction is exactly equal to the enthalpy change, of the reacting system.

So, the correct answer is option A.

Statement I : An exothermic reaction has a positive $\displaystyle \Delta H$ value
Because
Statement II : Heat must be added to the reaction for the reaction to occur

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1 .

  2. Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

  3. Statement 1 is correct but Statement 2 is not correct.

  4. Statement 1 is not correct but Statement 2 is correct.

  5. Both the Statement 1 and Statement 2 are not correct.


Correct Option: E
Explanation:
Answer (E)

An exothermic reaction has a negative ΔHΔH value,so statement 1 is not correct.
It is not necessary that heat must be added to the reaction for the reaction to occur, only endothermic reactions requires heat.
so both statement 1 and 2 are not correct.
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