K_p = 0.0198 at 721 K for the reaction 2HI (g) H₂ (g) + I₂ (g)
In a particular experiment, the partial pressures of H₂ and I₂ at equilibrium are 0.836 and 0.701 atm, respectively. The partial pressure of HI is __________ atm.

Question

K_p = 0.0198 at 721 K for the reaction 2HI (g) Kp = 0.0198 at 721 K for the reaction 2HI (g) H2 (g) + I2 (g) In a particular experiment, the partial pressures of H2 and I2 at equilibrium are 0.836 and 0.701 atm, respectively. The partial pressure of HI is __________ atm. A) 7.87 B) 29.6 C) 5.44 D) 0.108 E) 0.0116

K<sub>p</sub> = 0.0198 at 721 K for the reaction 2HI (g) H<sub>2</sub> (g) + I<sub>2</sub> (g) In a particular experiment, the partial pressures of H<sub>2</sub> and I<sub>2</sub> at equilibrium are 0.836 and 0.701 atm, respectively. The partial pressure of HI is __________ atm. A) 7.87 B) 29.6 C) 5.44 D) 0.108 E) 0.0116

H₂ (g) + I₂ (g)
In a particular experiment, the partial pressures of H₂ and I₂ at equilibrium are 0.836 and 0.701 atm, respectively. The partial pressure of HI is __________ atm.

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:
Kp = (PH2S * PI2S) / P2HI^2

We are given Kp = 0.0198 and PH2S = 0.836 atm, PI2S = 0.701 atm.

Thus,
0.0198 = (0.836 * 0.701) / P2HI^2
P2HI^2 = (0.836 * 0.701) / 0.0198
P2HI^2 = 29.6
P2HI = sqrt(29.6)
P2HI = 5.44 atm

Therefore, the partial pressure of HI is 5.44 atm. Hence, the correct answer is C.