Test Index

NCERT Class XI Chemistry Thermodynamics Solutions

© examsnet.com
Question : 12 of 22
Marks: +1, -0
Enthalpies of formation of CO(g),CO2(g),N2O(g)\mathrm{CO}_{(g)}, \mathrm{CO}_{2(g)}, \mathrm{N_2O}_{(g)} and N2O4(g)\mathrm{N_2O}_{4(g)} are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of DrH for the reaction :
N2O4(g)+3CO(g)\mathrm{N_2O}_{4(g)} + 3\mathrm{CO}_{(g)} → N2O(g)+3CO2(g)\mathrm{N_2O}_{(g)} + 3\mathrm{CO}_{2(g)}
Solution:  
N2O4(g)+3CO(g)\mathrm{N_2O}_{4(g)} + 3\mathrm{CO}_{(g)} → N2O(g)+3CO2(g)\mathrm{N_2O}_{(g)} + 3\mathrm{CO}_{2(g)}
ΔrH\Delta_r H = ΔfH[N2O(g)]+3ΔfH[CO2(g)]−(ΔfH[N2O4(g)]+3ΔrH[CO(g)])\Delta_f H[\mathrm{N_2O}_{(g)}] + 3\Delta_f H[\mathrm{CO}_{2(g)}] - (\Delta_f H[\mathrm{N_2O}_{4(g)}] + 3\Delta_r H[\mathrm{CO}_{(g)}])
= 81 + 3 × (–393) – {9.7 + 3(–110)} = 81 – 1179 – 9.7 + 330 = –777.7 kJmol−1\mathrm{kJ mol}^{-1}
© examsnet.com
Go to Question:
Prev Question
Next Question