Calculate K_pfor the reaction $\mathrm { H } _ { 2 } ( g ) + \mathrm { I } _ { 2 } ( g ) f \quad 2 \mathrm { HI } ( g )$ . Given that the mixture of H₂ and I₂ was maintained at 740 K until the system reached equilibrium, and the equilibrium mixture contained 3.68 × 10^?2 M HI, 8.31 × 10^?3 M H₂, and 7.72 × 10^?4 M I₂.

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:$$K_p = \frac{P_{HI}^2}{P_{H_2}P_{I_2}}$$where P is the partial pressure of each gas.At equilibrium, the partial pressures of the gases are:$$P_{HI} = 3.68 imes 10^{-2} ext{ atm}$$$$P_{H_2} = 8.31 imes 10^{-3} ext{ atm}$$$$P_{I_2} = 7.72 imes 10^{-4} ext{ atm}$$Substituting these values into the equilibrium constant expression, we get:$$K_p = \frac{(3.68 imes 10^{-2})^2}{(8.31 imes 10^{-3})(7.72 imes 10^{-4})} = 2.11 imes 10^2$$Therefore, the correct answer is B) 2.11 × 10².

Calculate Kp for the Reaction H2(g)+I2(g)f2HI(g)\mathrm { H } _ { 2 } ( g ) + \mathrm { I } _ { 2 } ( g ) f \quad 2 \mathrm { HI } ( g )H2​(g)+I2​(g)f2HI(g)

Calculate K_pfor the Reaction $\mathrm { H } _ { 2 } ( g ) + \mathrm { I } _ { 2 } ( g ) f \quad 2 \mathrm { HI } ( g )$