Question

At \(T\) K, the equilibrium constant for the reaction \[ {H}_2(g) + {Br}_2(g) \rightleftharpoons 2 {HBr}(g) \] is 1.6 \(\times\) 10\(^{1}\). If 10 bar of HBr is introduced into a sealed vessel at \(T\) K, the equilibrium pressure of HBr (in bar) is approximately:

• A. \(10.20\)
• B. \(10.95\)
• C. \(9.95\)
• D. \(11.95\)

Hint:

In equilibrium problems involving changes in concentration or pressure, always check if the changes make physical sense (e.g., pressure cannot be negative).

Answer 1

The Correct Option is C

Assuming ideal behavior and using the reaction quotient \(Q_p\), we start with initial pressures: \[ [{H}_2] = [{Br}_2] = 0, \quad [{HBr}] = 10 \, {bar} \] Let \(x\) be the change in pressure due to reaction at equilibrium: \[ [{H}_2] = x, \quad [{Br}_2] = x, \quad [{HBr}] = 10 - 2x \] The equilibrium constant \(K_p\) is given by: \[ K_p = \frac{[{HBr}]^2}{[{H}_2][{Br}_2]} = \frac{(10 - 2x)^2}{x^2} \] Setting \(K_p\) to 1.6 \(\times\) 10\(^{1}\) and solving for \(x\), we find: \[ 1.6 \times 10 = \frac{(10 - 2x)^2}{x^2} \] Solve this quadratic equation to find \(x\) and then use it to calculate the equilibrium pressure of HBr.