1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy of formation of CO(g), CO2(g), N2O(g) and N2O4(g) is -110, -393, +811 and 10 kj/mol respectively. For the reaction, N2O4(g)+3CO(g)→ N2O(g)+3CO2(g).Δ Hr (kJ/mol) is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

AIIMSAIIMS 2011Thermodynamics

Solution:

$N _2 O _{4( g )}+3 CO _{( g )} \longrightarrow N _2 O _{( g )}+3 CO _{2( g )}$
$\Delta H_{\text {reaction }}=\sum_{\text {Heat of formation of products }}-\sum_{\text {Heat of formation of reactants }}$
$\Delta H_{\text {reaction }}=\left[\Delta H_f N_2 O+3 \times \Delta H_f C O_2\right]-\left[\Delta H_f N_2 O_4+3 \times \Delta H_f C O\right]$
$\Delta H_r=[+811+3(-393)]-[10+3(-110)]$
$=[811-1179]-[-320]=-368+320$
$=-48 kJ / mol$