Question

# Given,C(graphite)+O2(g)→CO2(g);ΔrHo=−393.5kJmol−1H2(g)+12O2(g)→H2O(l);ΔrHo=−285.8kJmol−1CO2(g)+2H2O(l)→CH4(g)+2O2(g);ΔrHo=+890.3kJmol−1Based on the above thermochemical equations, the value of ΔrHo at 298K for the reactionC(graphite)+2H2(g)→CH4(g) will be:

A
+144.0kJmol1
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B
74.8kJmol1
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C
144.0kJmol1
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D
+78.8kJmol1
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Solution

## The correct option is C −74.8kJmol−1Cgraphite+O2(g)→CO2(g);ΔrH∘=−393.5 .....(1)H2(g)+12O2(g)→H2O(l);ΔrH∘=−285.8.....(2)CO2(g)+2H2O(l)→CH4(g)+2O2(g);ΔrH∘=+890.3......(3)Add reactions (1) and (3)C(graphite)+2H2O(l)→CH4(g)+O2(g) ……(4)ΔH=−393.5+890.3=496.8 kJ/molMultiply reaction (2) with 22H2(g)+O2(g)→2H2O(l) …..(5)ΔH=−285.8×2=−571.6 kJ/molAdd reactions (4) and (5)C(graphite)+2H2(g)→CH4(g)ΔH=496.8−571.6=−74.8 kJ/mol

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