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Question

Given,
C(graphite)+O2(g)CO2(g);ΔrHo=393.5kJmol1

H2(g)+12O2(g)H2O(l);ΔrHo=285.8kJmol1
CO2(g)+2H2O(l)CH4(g)+2O2(g);ΔrHo=+890.3kJmol1
Based on the above thermochemical equations, the value of ΔrHo at 298K for the reaction
C(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 74.8kJmol1
Cgraphite+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/mol
Multiply reaction (2) with 2
2H2(g)+O2(g)2H2O(l) …..(5)
ΔH=285.8×2=571.6 kJ/mol
Add reactions (4) and (5)
C(graphite)+2H2(g)CH4(g)
ΔH=496.8571.6=74.8 kJ/mol

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