Question

Consider the reaction.

2 A(g) = B(g)
Kp = 7.10 x 10-5 at 500 K
If a sample of A(g) at 2.70 bar is heated to 500 K, what is the pressure of B(g) at equilibrium?

Answer 1

The pressure of B(g) at equilibrium can be calculated using the equilibrium constant, Kp, and the initial pressure of A(g). By rearranging the equation, we can solve for [B] and convert it back to pressure. In this case, the pressure of B(g) at equilibrium is approximately 0.01906 bar.

Step-by-step explanation:

The equilibrium constant, Kp, relates to the ratio of product partial pressures to reactant partial pressures at equilibrium. In this case, the reaction is 2 A(g) = B(g) and the given value of Kp is 7.10 x 10-5 at 500 K. To find the pressure of B(g) at equilibrium, we can use the equation Kp = ([B]2 / [A]2). Assuming the initial pressure of A(g) is 2.70 bar, we can substitute this value into the equation, solve for [B], and then convert it back to pressure.

Starting with Kp = ([B]2 / [A]2), we can rearrange the equation to solve for [B]: [B] = sqrt(Kp * [A]2). Plugging in the values, [A] = 2.70 bar and Kp = 7.10 x 10-5, we get [B] = sqrt((7.10 x 10-5) * (2.702)). Simplifying, we find [B] ≈ 0.01906 bar. Therefore, the pressure of B(g) at equilibrium is approximately 0.01906 bar.