Question

The enthalpy of formation of $CO_{\left(g\right),} CO_{2\left(g\right)}, N_{2}O_{\left(g\right)}$ and $N_{2}O_{4\left(g\right)}$ is -110, -393, +811 and 10 kj/mol respectively. For the reaction, $N_{2}O_{4\left(g\right)}+3CO_{\left(g\right)}\to N_{2}O_{\left(g\right)}+3CO_{2\left(g\right)}.\Delta H_{r} \left(kJ/mol\right)$ is

• A. -212
• B. 212
• C. 48
• D. -48

Answer 1

Answer: (D)

-48

Answer 2

${N2O4_{(g)} + 3CO_{(g)} ->N2O_{(g)} + 3CO2_{(g)}}$ $\Delta H_{\text{reaction} }= \sum_{\text{Heat of formation of products} }-\sum_{\text{Heat of formation of reactants}}$ $\Delta H_{\text{reaction} }=\left[\Delta H_{f} N_{2}O+3\times\Delta H_{f} CO_{2}\right]-\left[\Delta H_{f} N_{2}O_{4}+3\times \Delta H_{f} CO\right]$ $\Delta H_{r}=\left[+811+3\left(-393\right)\right]-\left[10+3\left(-110\right)\right]$ $=\left[811-1179\right]-\left[-320\right]=-368+320$ $=-48\, kJ/mol$