    Question

# Given, C(graphites)+O2(g)→CO2(g) Δr H∘=−393.5 kJ mol−1 H2(g)+12 O2(g)→H2O(l); ΔrH∘=−285.8 kJ mol−2 CO2(g+2H2O(l)→CH4(g)+2O2(g) ΔrH∘=+890.3 kJ mol−1 Based on the above thermochemical equations, the value of Δr H∘ at 298K for the reaction, C(graphite)+2H2(g)→CH4(g) will be

A
+78.8kJmol(1)
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B
+144.0kJmol(1)
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C
74.8kJmol(1)
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D
144.0kJmol(1)
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Solution

## The correct option is B −74.8kJmol(−1)Based on given Δr H∘ Δf H∘=H∘CO2=−393.5kJmol−1 . . . . (i) ΔfH∘=H∘(H2O)=−285.8kJmol−1 . . . . (ii) ΔfH∘=H∘O2=0.00(elements) Required thermal reaction is for ΔfH∘ of (CH4) Thus, from III 890.3=[ΔfH∘(CH4)+2 Δf H∘(O2)] −[ΔfH∘(CO2)+2 ΔfH∘(H2O)] =Δf H∘(CH4)+0−[−393.5−2×285.5] ∴ ΔfH∘(CH4)=−74.8 kJ/mol  Suggest Corrections  0      Similar questions  Related Videos   Thermochemistry
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