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Question: What will be the standard internal energy change for the reaction at 298 K? \[OF_{2(g)} + H_{2}O_{(...

What will be the standard internal energy change for the reaction at 298 K?

OF2(g)+H2O(g)→O2(g)+2HF(g);ΔHo=−310kJOF_{2(g)} + H_{2}O_{(g)} \rightarrow O_{2(g)} + 2HF_{(g);}\Delta H^{o} = - 310kJ

A

−312.5kJ- 312.5kJ

B

−125.03kJ- 125.03kJ

C

−310kJ- 310kJ

D

−156kJ- 156kJ

Answer

−312.5kJ- 312.5kJ

Explanation

Solution

: ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_{g}RT

ΔH=−310×103J,Δng=3−2=1,R=8.314JK−1mol−1\Delta H = - 310 \times 10^{3}J,\Delta n_{g} = 3 - 2 = 1,R = 8.314JK^{- 1}mol^{- 1}

T = 298 K

ΔU=−310×103−(1×8.314×298)\Delta U = - 310 \times 10^{3} - (1 \times 8.314 \times 298)

=−312477×103J=−312.5kJ= - 312477 \times 10^{3}J = - 312.5kJ