Heat of formation of water is -68.3K.cal, heat of combustion of acetylene and ethylene are -310.6, -337.2 K.cal respectively, calculate the heat of reaction for the hydrogenation of acetylene at constant volume, at 25°C in K.cal

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-41.1

-32.2

-56.7

-22.4

answer is A.

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Detailed Solution

H2+12O2→H2O;ΔH1=-68.3K.calC2H2+52O2→2CO2+H2O;ΔH2=-310.6K.calC2H4+3O2→2CO2+2H2O;ΔH3=-337.2K.calC2H2+H2→C2H4;ΔH=?ΔH=ΔH1+ΔH2-ΔH3∴C2H2(g)+H2(g)→C2H4(g);ΔH=-41.7K.CalΔH=ΔU+ΔnRTΔn=1-2=-1;R=2cal,T=298 KΔU=-41.7+1×21000×298 K=-41.1K.Cal

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