Question:
The value of \(\Delta_r H\) for the reaction:
\[
{N}_2{O}_4(g) + 3 {CO}(g) \rightarrow {N}_2{O}(g) + 3 {CO}_2(g)
\]
(in kJ) is:
(Given: Enthalpies of formation in kJ mol\(^{-1}\) are \({CO}(g) = -110\), \({CO}_2(g) = -393\), \({N}_2{O}(g) = 81\), and \({N}_2{O}_4(g) = 9.7\))
The value of \(\Delta_r H\) for the reaction:
\[ {N}_2{O}_4(g) + 3 {CO}(g) \rightarrow {N}_2{O}(g) + 3 {CO}_2(g) \] (in kJ) is:
(Given: Enthalpies of formation in kJ mol\(^{-1}\) are \({CO}(g) = -110\), \({CO}_2(g) = -393\), \({N}_2{O}(g) = 81\), and \({N}_2{O}_4(g) = 9.7\))
\[ {N}_2{O}_4(g) + 3 {CO}(g) \rightarrow {N}_2{O}(g) + 3 {CO}_2(g) \] (in kJ) is:
(Given: Enthalpies of formation in kJ mol\(^{-1}\) are \({CO}(g) = -110\), \({CO}_2(g) = -393\), \({N}_2{O}(g) = 81\), and \({N}_2{O}_4(g) = 9.7\))
Show Hint
Use enthalpies of formation and Hess's law to calculate reaction enthalpy.
Updated On: Jun 2, 2025
- +678
- -678
- -778
- +578
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The Correct Option is C
Solution and Explanation
\[
\Delta_r H = \sum \Delta H_f ({products}) - \sum \Delta H_f ({reactants})
\]
Calculate sum of products:
\[
\Delta H_f ({N}_2{O}) + 3 \times \Delta H_f ({CO}_2) = 81 + 3 \times (-393) = 81 - 1179 = -1098
\]
Calculate sum of reactants:
\[
\Delta H_f ({N}_2{O}_4) + 3 \times \Delta H_f ({CO}) = 9.7 + 3 \times (-110) = 9.7 - 330 = -320.3
\]
Calculate reaction enthalpy:
\[
\Delta_r H = -1098 - (-320.3) = -1098 + 320.3 = -777.7 \approx -778 \, {kJ}
\]
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