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Question: For a reaction, \(2K_{(g)} + L_{(g)} \rightarrow 2M_{(g);}\Delta U^{o} = - 10.5kJ\)and \(\Delta S^{o...

For a reaction, 2K(g)+L(g)→2M(g);ΔUo=−10.5kJ2K_{(g)} + L_{(g)} \rightarrow 2M_{(g);}\Delta U^{o} = - 10.5kJand ΔSo=−44.1JK−1\Delta S^{o} = - 44.1JK^{- 1} Calculate ΔGo\Delta G^{o} for the reaction and predict whether the reaction will be spontaneous or non – spontaneous?

A

ΔG=+0.16kJ,\Delta G = + 0.16kJ, non – spontaneous

B

ΔG=−0.16kJ,\Delta G = - 0.16kJ, spontaneous

C

ΔG=+26.12kJ,\Delta G = + 26.12kJ, non – spontaneous

D

ΔG=−26.12kJ,\Delta G = - 26.12kJ, spontaneous

Answer

ΔG=+0.16kJ,\Delta G = + 0.16kJ, non – spontaneous

Explanation

Solution

: 2K+L→2M2K + L \rightarrow 2M

Δng=2−3=−1\Delta n_{g} = 2 - 3 = - 1

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

=−10.5×103+(−1×8.314×298)= - 10.5 \times 10^{3} + ( - 1 \times 8.314 \times 298)

=−10500+(−2477.572)=−12977.57J=−12.98kJΔGo=ΔHo−TΔSo= - 10500 + ( - 2477.572) = - 12977.57J = - 12.98kJ\Delta G^{o} = \Delta H^{o} - T\Delta S^{o}

=−12.98−298(−44.1×10−3)= - 12.98 - 298( - 44.1 \times 10^{- 3})

=−12.98+13.14=0.16kJ= - 12.98 + 13.14 = 0.16kJ

Since ΔGo\Delta G^{o}is ve hence it is non – spontaneous.