Q.

The value of enthalpy change (ΔH) for the reaction C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l) at 27 °C is 1366.5kJmol1. The value of internal energy change for the above reaction at this temperature will be

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a

1369.0kJmol1

b

1361.5kJmol1

c

1371.5kJmol1

d

1364.0kJmol1

answer is C.

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Detailed Solution

Since ΔrH=ΔrU+ΔvgRT, we have ΔrU=ΔrHΔvgRT

For the given reaction Δvg=1. Hence

ΔrU=1366.5kJmol1(1)8.314×103kJK1mol1(300K)=(1366.5+2.494)kJmol11364.0kJmol1

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