Consider the reaction:

\[ 2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g), \quad K_p = 28.4 \text{ at } 298K \]

In a reaction mixture at equilibrium, the partial pressure of NO is 114 torr and that of Br₂ is 137 torr. What is the partial pressure of NOBr?

a) 98.2 torr
b) 103.8 torr
c) 137 torr
d) 171.2 torr

To calculate the equilibrium partial pressure of NOBr, we use the equilibrium expression together with the provided Kp value and the known partial pressures of NO and Br2. The correct partial pressure of NOBr is found to be d) 171.2 torr.

Explanation:

To determine the partial pressure of NOBr at equilibrium for the reaction 2NO(g) + Br2(g) \<=> 2NOBr(g), with a given equilibrium constant (Kp = 28.4 at 298 K) and provided partial pressures for NO (114 torr) and Br2 (137 torr), we will use the equilibrium expression associated with the reaction. The equilibrium expression in terms of partial pressures is given by Kp = (P_NOBr)^2 / (P_NO)^2(P_Br2).

Given the Kp value, we can solve for the unknown partial pressure of NOBr (P_NOBr) by rearranging the formula and substituting the known values: Kp * (P_NO)^2(P_Br2) = (P_NOBr)^2. After substituting the given pressures and solving the resulting quadratic equation, we can find the equilibrium partial pressure of NOBr. Since P_NO and P_Br2 are given in torr, P_NOBr will also be expressed in torr.