Tag: enthalpy changes and enthalpy profile diagrams

Questions Related to enthalpy changes and enthalpy profile diagrams

Enthalpy of the system is given as

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $\,H + PV$

  2. $\,U + PV$

  3. $\,U - PV$

  4. $\,H - PV$

Show answer
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?

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

Show answer
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.

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

Show answer
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, $\Delta H > \Delta U$?

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

Show answer
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:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $U + PV$

  2. $H = PV$

  3. $U - PV$

  4. $ H - V$

Show answer
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. $\Delta H = \Delta U - RT$

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

Show answer
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 $\Delta U$.

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

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

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

  4. None of these

Show answer
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:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  1. $2RT$

  2. $-2RT$

  3. $3RT$

  4. $RT$

Show answer
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:

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

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

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

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

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

Show answer
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

chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

  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.

Show answer
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.