Nitrosyl bromide decomposes according to the chemical equation below. 2 NOBr(g) $\rightleftharpoons$ 2 NO(g) + Br₂(g)
When 0.260 atm of NOBr is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr decomposes. What is the equilibrium constant, K_p, for the reaction?

Question

Nitrosyl bromide decomposes according to the chemical equation below. 2 NOBr(g)

\rightleftharpoons

2 NO(g) + Br₂(g)
When 0.260 atm of NOBr is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr decomposes. What is the equilibrium constant, K_p, for the reaction?

Quizplus · Accepted Answer

The equilibrium constant, K_p, is given by the expression:K_p = (P_NO)²(P_Br2) / (P_NOBr)²At equilibrium, 22% of the NOBr decomposes, so the partial pressure of NOBr is:P_NOBr = 0.260 atm * (1 - 0.22) = 0.203 atmThe partial pressure of NO and Br₂ can be calculated using the stoichiometry of the reaction:P_NO = P_Br2 = 0.260 atm * 0.22 / 2 = 0.0286 atmSubstituting these values into the expression for K_p, we get:K_p = (0.0286 atm)² / (0.203 atm)² = 2.3 × 10^-3

Nitrosyl Bromide Decomposes According to the Chemical Equation Below ⇌\rightleftharpoons⇌ 2 NO(g) + Br2(g) When 0

Nitrosyl Bromide Decomposes According to the Chemical Equation Below $\rightleftharpoons$ 2 NO(g) + Br₂(g)
When 0