The molar enthalpies of combusion of C2H2(g),C(graphite) and H2(g)are -1300,-394,-286 kJmol−1respectively. The standard enthalpy of formation of C2H2(g)is
A
-226 kJmol−1
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B
-626 kJmol−1
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C
226 kJmol−1
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D
626 kJmol−1
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Solution
The correct option is D 626 kJmol−1 (i) C (graphite) +O2→CO2−394 (ii) H2+12O2→H2O−286 (iii) C2H2+52O2→2CO2+H2O−1300 2 x (i) + (ii) + (iii) 2 C (graphite) +H2(g)→C2H2 ΔH=−2×394−286+1300 ΔH=226KJ mol−1