Q.

For the reaction  H2F2(g)H2(g)+F2(g)

ΔU=59.6kJ  mol1  at  270C

The enthalpy change for the above reaction is (-)_____  kJmol1 [nearest integer] Given  R=8.314  JK1  mol1

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answer is 57.

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Detailed Solution

ΔH=ΔU+ΔngRT Δng=nR(g)np(g)=21=1 ΔH=59.6+1×8.314×300×103=57.10

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For the reaction  H2F2(g)→H2(g)+F2(g)ΔU=−59.6 kJ  mol−1  at  270CThe enthalpy change for the above reaction is (-)_____  kJ mol−1 [nearest integer] Given  R=8.314  JK−1  mol−1