Question

Determine $\Delta {\text{H}}$for the following reaction at $500\,{\text{K}}$and constant pressure:
${\text{CO}}\,{\text{(g)}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{O(g)}} \to \,{\text{C}}{{\text{O}}_2}\,{\text{(g)}}\,{\text{ + }}\,{{\text{H}}_{\text{2}}}{\text{(g)}}$

Use the following data:

Substance	${{\text{C}}_{\text{p}}}$(J/mol K)	${\Delta _{\text{f}}}{\text{H}}$($298\,{\text{K}}$) (kJ/mol)
CO	$29.12$	$\, - 110.5$
${{\text{H}}_{\text{2}}}{\text{O}}$	$33.58$	$ - 241.8$
${\text{C}}{{\text{O}}_{\text{2}}}$	$37.11$	$ - 393.5$
${{\text{H}}_{\text{2}}}$	$29.89$	$\,0.0$

A. $\Delta {\text{H}}\, = \, - {\text{30}}{\text{.3}}\,{\text{kJ}}$
B. $\Delta {\text{H}} = \, - 5{\text{0}}{\text{.3}}\,{\text{kJ}}$
C. $\Delta {\text{H}}\, = \, - 4{\text{0}}{\text{.3}}\,{\text{kJ}}$
D. $\Delta {\text{H}}\, = \, - 2{\text{0}}{\text{.3}}\,{\text{kJ}}$

Answer 1

Hint: Molar heat capacity is defined as the heat required to increase the temperature of one mole of a substance by$1{\,^ \circ }{\text{C}}$ . If the same reaction occurs at two different temp and constant pressure and heat change at a temperature is given then we can determine the heat change at another temperature by using Kirchhoff’s equation.

Formula: $\Delta {\text{H}} = \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$

Complete step-by-step answer:Kirchhoff’s equation relates heat change at two different temperatures.
The Kirchhoff’s equation is as follows:
$\Delta {\text{H}} = \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$ 
Where,
$\Delta {\text{H}}$is the change in heat
$\Delta {{\text{C}}_{\text{p}}}$ is the molar heat capacity at constant pressure
${{\text{T}}_2}$is the final temperature
${{\text{T}}_1}$is the initial temperature
We will substitute the enthalpy at both temperature and rearrange the equation for enthalpy change at final temperature as follows:
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) - \Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right) = \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right)\, + \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$…..$(1)$
We will use the given molar heat capacities at constant P and $298\,{\text{K}}$to determine the change in heat capacity as follows:
The formula to determine the change in heat capacity as follows:
$\Delta {{\text{C}}_{\text{p}}} = \,\sum {{{\text{C}}_{\text{p}}}({\text{products)}}} \, - \,\sum {{{\text{C}}_{\text{p}}}({\text{reactants)}}} $
Where,
$\sum {{{\text{C}}_{\text{p}}}({\text{products)}}} \,$is the summation of heat capacities of products
$\sum {{{\text{C}}_{\text{p}}}({\text{reactants)}}} $is the summation of heat capacities of products
For the given reaction$\Delta {{\text{C}}_{\text{p}}}$formula can be written as follows:
$\Delta {{\text{C}}_{\text{p}}} = \,\left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{C}}{{\text{O}}_{\text{2}}}{\text{)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({{\text{H}}_{\text{2}}}{\text{)}}} \right] - \left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{CO)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({{\text{H}}_{\text{2}}}{\text{O)}}} \right]$
On substituting the values of molar heat capacities,
$\Delta {{\text{C}}_{\text{p}}} = \,\left[ {1\, \times {{\text{C}}_{\text{p}}}(33.11{\text{)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({\text{29}}{\text{.89)}}} \right] - \left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{29}}{\text{.12)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({\text{33}}{\text{.58)}}} \right]$
\[\Delta {{\text{C}}_{\text{p}}} = \,67 - 62.7\]
\[\Delta {{\text{C}}_{\text{p}}} = \,4.3\,{\text{J}}\]
Convert the \[\Delta {{\text{C}}_{\text{p}}}\]from joule to kJ as follows:
\[1000\,{\text{J}}\,{\text{ = }}\,{\text{1}}\,{\text{kJ}}\]
\[4.3\,{\text{J = }}\,0.0043\,{\text{kJ}}\]
We will use the given molar heat capacities at constant P and $298\,{\text{K}}$to determine the change in enthalpy as follows:
The formula to determine the change in enthalpy as follows:
$\Delta {{\text{H}}_1} = \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}} $
Where,
$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \,$is the summation of enthalpy of products
$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}} $is the summation of enthalpy of products
For the given reaction$\Delta {{\text{H}}_1}$formula can be written as follows:
$\Delta {{\text{H}}_1} = \,\left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{C}}{{\text{O}}_{\text{2}}}{\text{)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{H}}_{\text{2}}}{\text{)}}} \right] - \left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{CO)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{H}}_{\text{2}}}{\text{O)}}} \right]$
On substituting the values of ${\Delta _{\text{f}}}{\text{H}}$,
${\Delta _{\text{f}}}{\text{H}} = \,\left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}( - 393.5{\text{)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}(0.0{\text{)}}} \right] - \left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}({\text{ - 110}}{\text{.5)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}({\text{ - 241}}{\text{.8)}}} \right]$
\[{\Delta _{\text{f}}}{\text{H}} = \, - 393.5\, + \,352.3\]
\[{\Delta _{\text{f}}}{\text{H}} = \, - 41.2\,\,{\text{kJ}}\]
Now substitute $ - 41.2$kJ for$\Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right)$, \[{\text{0}}{\text{.0043}}\,{\text{kJ}}\] for \[\Delta {{\text{C}}_{\text{p}}}\], $500\,{\text{K}}$for ${{\text{T}}_2}$and $298\,{\text{K}}$for ${{\text{T}}_1}$in equation$(1)$.
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\left( { - 41.2} \right)\, + \,\left( {{\text{0}}{\text{.0043}}} \right)\left( {500 - {\text{298}}} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\left( { - 41.2} \right)\, + \,\left( {0.8686} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \, - 40.3\,{\text{kJ}}$
So, the enthalpy change at $500\,{\text{K}}$is $\, - 40.3\,{\text{kJ}}$.
Kirchhoff’s equation relates heat change at two different temperatures.
The Kirchhoff’s equation is as follows:$\Delta {\text{H}} = \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$ 
Where,
$\Delta {\text{H}}$is the change in heat
$\Delta {{\text{C}}_{\text{p}}}$ is the molar heat capacity at constant pressure
${{\text{T}}_2}$is the final temperature
${{\text{T}}_1}$is the initial temperature
We will substitute the enthalpy at both temperature and rearrange the equation for enthalpy change at final temperature as follows:
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) - \Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right) = \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right)\, + \,\Delta {{\text{C}}_{\text{p}}}\left( {{{\text{T}}_2} - {{\text{T}}_{\text{1}}}} \right)$…..$(1)$
We will use the given molar heat capacities at constant P and $298\,{\text{K}}$to determine the change in heat capacity as follows:
The formula to determine the change in heat capacity as follows:
$\Delta {{\text{C}}_{\text{p}}} = \,\sum {{{\text{C}}_{\text{p}}}({\text{products)}}} \, - \,\sum {{{\text{C}}_{\text{p}}}({\text{reactants)}}} $
Where,
$\sum {{{\text{C}}_{\text{p}}}({\text{products)}}} \,$is the summation of heat capacities of products
$\sum {{{\text{C}}_{\text{p}}}({\text{reactants)}}} $is the summation of heat capacities of products
For the given reaction$\Delta {{\text{C}}_{\text{p}}}$formula can be written as follows:
$\Delta {{\text{C}}_{\text{p}}} = \,\left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{C}}{{\text{O}}_{\text{2}}}{\text{)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({{\text{H}}_{\text{2}}}{\text{)}}} \right] - \left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{CO)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({{\text{H}}_{\text{2}}}{\text{O)}}} \right]$
On substituting the values of molar heat capacities,
$\Delta {{\text{C}}_{\text{p}}} = \,\left[ {1\, \times {{\text{C}}_{\text{p}}}(33.11{\text{)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({\text{29}}{\text{.89)}}} \right] - \left[ {1\, \times {{\text{C}}_{\text{p}}}({\text{29}}{\text{.12)}} + \,1\,\, \times {{\text{C}}_{\text{p}}}({\text{33}}{\text{.58)}}} \right]$
\[\Delta {{\text{C}}_{\text{p}}} = \,67 - 62.7\]
\[\Delta {{\text{C}}_{\text{p}}} = \,4.3\,{\text{J}}\]
Convert the \[\Delta {{\text{C}}_{\text{p}}}\]from joule to kJ as follows:
\[1000\,{\text{J}}\,{\text{ = }}\,{\text{1}}\,{\text{kJ}}\]
\[4.3\,{\text{J = }}\,0.0043\,{\text{kJ}}\]
We will use the given molar heat capacities at constant P and $298\,{\text{K}}$to determine the change in enthalpy as follows:
The formula to determine the change in enthalpy as follows:
$\Delta {{\text{H}}_1} = \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \, - \,\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}} $
Where,
$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{products)}}} \,$is the summation of enthalpy of products
$\sum {{\Delta _{\text{f}}}{\text{H}}\,({\text{reactants)}}} $is the summation of enthalpy of products
For the given reaction$\Delta {{\text{H}}_1}$formula can be written as follows:
$\Delta {{\text{H}}_1} = \,\left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{C}}{{\text{O}}_{\text{2}}}{\text{)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{H}}_{\text{2}}}{\text{)}}} \right] - \left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}\,({\text{CO)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}\,({{\text{H}}_{\text{2}}}{\text{O)}}} \right]$
On substituting the values of ${\Delta _{\text{f}}}{\text{H}}$,
${\Delta _{\text{f}}}{\text{H}} = \,\left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}( - 393.5{\text{)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}(0.0{\text{)}}} \right] - \left[ {1\, \times {\Delta _{\text{f}}}{\text{H}}({\text{ - 110}}{\text{.5)}} + \,1\,\, \times {\Delta _{\text{f}}}{\text{H}}({\text{ - 241}}{\text{.8)}}} \right]$
\[{\Delta _{\text{f}}}{\text{H}} = \, - 393.5\, + \,352.3\]
\[{\Delta _{\text{f}}}{\text{H}} = \, - 41.2\,\,{\text{kJ}}\]
Now substitute $ - 41.2$kJ for$\Delta {{\text{H}}_1}\left( {{{\text{T}}_1}} \right)$, \[{\text{0}}{\text{.0043}}\,{\text{kJ}}\] for \[\Delta {{\text{C}}_{\text{p}}}\], $500\,{\text{K}}$for ${{\text{T}}_2}$and $298\,{\text{K}}$for ${{\text{T}}_1}$in equation$(1)$.
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\left( { - 41.2} \right)\, + \,\left( 7{{\text{0}}{\text{.0043}}} \right)\left( {500 - {\text{298}}} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \,\left( { - 41.2} \right)\, + \,\left( {0.8686} \right)$
$\Delta {{\text{H}}_2}\left( {{{\text{T}}_2}} \right) = \, - 40.3\,{\text{kJ}}$
So, the enthalpy change at $500\,{\text{K}}$is $\, - 40.3\,{\text{kJ}}$.

Therefore, option (C) $\Delta {\text{H}}\, = \, - 4{\text{0}}{\text{.3}}\,{\text{kJ}}$ is correct.therefore, option (C) $\Delta {\text{H}}\, = \, - 4{\text{0}}{\text{.3}}\,{\text{kJ}}$ is correct.

Note:To determine the change in molar heat capacity a balanced chemical equation is necessary. According to Kirchhoff's equation, heat capacity varies with temperature. On increasing the temperature heat capacity increases. The Heat capacity in terms of mole is an intensive property and it is independent of quantity of substance. It gives information on whether the work can be done or not by the system also.