What happens to the following reaction at equilibrium if the pressure is decreased?

2H2(g) + O2(g) 2H2O(g)

A. The equilibrium shifts left because Q < K.
B. The equilibrium shifts right because Q < K.
C. The equilibrium shifts right because Q > K.
D. The equilibrium shifts left because Q > K.

K = (H2O)^2/(H2)^2(O2)

In increase in P means the reaction will shift to the side with fewer mols. So increase in P shifts to the right; therefore, a decrease must shift to the left. There are two answers for left (A and D). Can you figures with Q is larger or smaller.

To determine what happens to the reaction at equilibrium when the pressure is decreased, we need to consider Le Chatelier's principle. According to this principle, if a system at equilibrium is subjected to a change in pressure, it will respond in a way that minimizes the effect of that change.

In this case, decreasing the pressure will cause the system to respond by trying to increase the pressure again. One way to achieve this is by shifting the equilibrium in the direction that produces fewer moles of gas.

Looking at the balanced equation:
2H2(g) + O2(g) -> 2H2O(g)

The left side of the equation has fewer gas molecules (2) compared to the right side (2 + 1 = 3). Therefore, to decrease the pressure, the system will shift to the right, where fewer moles of gas are present.

Therefore, the correct answer is:

C. The equilibrium shifts right because Q > K.