Two moles of an ideal solution is prepared by mixing pure liquids ‘A’ and ‘B’ at $$27$$ºC. The mole percent of ‘A’ in the vapor above the solution is $$60$$. Vapor pressure of pure liquids A & B are $$PA0=120mm$$ Hg and $$PB0=80mm$$ Hg respectively. What is the magnitude of free energy change of mixing ΔmixGfor solution in calories? (Take$$, $$R=2$$ cal $$K-1$$ $$mol-1$$,$$log2=0.3$$,$$log3=0.48$$,$$log5=0.7$$,$$2.303$$≃$$2.3$$ $$)

Question

Two moles of an ideal solution is prepared by mixing pure liquids ‘A’ and ‘B’ at $$27$$ºC. The mole percent of ‘A’ in the vapor above the solution is $$60$$. Vapor pressure of pure liquids A & B are $$PA0=120mm$$ Hg and $$PB0=80mm$$ Hg respectively. What is the magnitude of free energy change of mixing ΔmixGfor solution in calories? (Take$$, $$R=2$$ cal $$K-1$$ $$mol-1$$,$$log2=0.3$$,$$log3=0.48$$,$$log5=0.7$$,$$2.303$$≃$$2.3$$ $$)

Infinity Learn · Accepted Answer

Δ$$mixG=$$ΔmixH−TΔmixSFor ideal solution, Δ$$mixH=0$$Δ$$mixG=$$−TΔmixSΔ$$mix=$$−$$2.303R$$×ntotal∑xilogxi≃−$$2.3$$×R×ntotal∑$$xilogxi1Ps=0.6120+0.480$$⇒$$Ps=100100=120$$×$$XAL+80$$×$$1$$−$$XA1XAl=12XBl=12$$Δ$$mixS=$$−$$2.3$$×$$2$$×$$212log12+12log12cal=2.76cal$$Δ$$mixG=$$−$$300$$×$$2.76cal=$$−$$828cal$$Δ$$mixG=828cal$$

Detailed Solution

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