The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below :$$CO(g)+H2O(g)$$⇌$$CO2(g)+H2(g)$$Δ$$H300K$$∘$$=$$−$$41.16kJmol$$−$$1$$Δ$$S300K$$∘$$=$$−$$4.24$$×$$10$$−$$2kJmol$$−$$1$$Δ$$H1200K$$∘$$=$$−$$32.93kJmol$$−$$1$$Δ$$S1200K$$∘$$=$$−$$2.96$$×$$10$$−$$2kJmol$$−$$1Calculate$$ Kp at each temperature and predict the direction of reaction at $$300$$ K and $$1200$$ K, when $$PCO=PCO2=PH2=PH2O=1$$ atm at initial state.

Question

The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below :$$CO(g)+H2O(g)$$⇌$$CO2(g)+H2(g)$$Δ$$H300K$$∘$$=$$−$$41.16kJmol$$−$$1$$Δ$$S300K$$∘$$=$$−$$4.24$$×$$10$$−$$2kJmol$$−$$1$$Δ$$H1200K$$∘$$=$$−$$32.93kJmol$$−$$1$$Δ$$S1200K$$∘$$=$$−$$2.96$$×$$10$$−$$2kJmol$$−$$1Calculate$$ Kp at each temperature and predict the direction of reaction at $$300$$ K and $$1200$$ K, when $$PCO=PCO2=PH2=PH2O=1$$ atm at initial state.

Infinity Learn · Accepted Answer

At equilibrium:    ΔG∘$$=$$ΔH∘−TΔS∘∴    Δ$$G300K$$∘$$=$$−$$41.16$$−$$300$$×−$$4.24$$×$$10$$−$$2=$$−$$28.44kJ$$ and     Δ$$G1200K$$∘$$=$$−$$32.93$$−$$1200$$×−$$2.96$$×$$10$$−$$2=+2.59kJThus$$, at $$300$$ K, reaction will proceed in forward direction and at $$1200$$ K in backward direction. Also, ΔG∘$$=$$−$$2.303RTlog$$⁡Kp At  $$300K$$,−$$28.44=$$−$$2.303$$×$$8.314$$×$$10$$−$$3$$×$$300log$$⁡Kp∴ $$Kp=8.94$$×$$104Similarly$$ , at $$1200$$ K, Kp $$=$$ $$0.77$$

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App