Given C_{\text{graphite}}+O_2(g)\toCO_2(g); \Delta_rH^\circ=-393.5kJmol^{-1} H_2(g)+1/2O_2(g)\toH_2O(I); \Delta_rH^\circ=-285.8kJmol^{-1} CO_2(g)+2H_2O(I)\toCH_4(g)+2O_2(g); \Delta_rH^\circ=+890.3kJmol^{-1} Based on the above thermochemical equations,the value of \Delta_rH^\circ at 298 K for the reaction C_{\text{graphite}}+2H_2(g)\toCH_4(g) will be :- [Main 2017]

Correct Answer is : −74.8kJmol−1

JEE Main 2017 Solved Paper Solved Paper

Section: Chemistry

Question : 77 of 90

Marks: +1, -0

Given
Cgraphite+O2(g)→CO2(g); ΔrH°=−393.5kJmol−1
H2(g)+

O2(g)→H2O(I); ΔrH°=−285.8kJmol−1
CO2(g)+2H2O(I)→CH4(g)+2O2(g); ΔrH°=+890.3kJmol−1
Based on the above thermochemical equations,the value of ΔrH° at 298 K for the reaction
Cgraphite+2H2(g)→CH4(g) will be :-

[Main 2017]

+74.8kJmol−1
+144.0kJmol−1
−74.8kJmol−1
−144.0kJmol−1

Validate

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