Consider the following reaction at equilibrium. What effect will reducing the pr

Question

Consider the following reaction at equilibrium. What effect will reducing the pressure have on the system?

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H° = -2220 kJ

Consider the following reaction at equilibrium. What effect will reducing the pressure have on the system?

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H° = -2220 kJ

A. The reaction will shift to the right in the direction of products. B. No effect will be observed. C. The equilibrium constant will decrease. D. The reaction will shift to the left in the direction of reactants. E. The equilibrium constant will increase.

Explanation / Answer

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H° = -2220 kJ

As the number of gas molecules is more on left side of the reaction, So, according to Le Chatlier's principle, reducing the pressure will shit the equilibrium to the left side increasint the number of molecules on the left side.

D. The reaction will shift to the left in the direction of reactants.

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