At 340 K, K_p = 69 for the reaction H₂(g)+ I₂(g) 2HI(g). 50.0 g of HI is injected into an evacuated 5.00-L rigid cylinder at 340 K. What is the total pressure inside the cylinder when the system comes to equilibrium?

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation: H₂(g) + I₂(g) ↔ 2HI(g) Next, we can use the equation: K_p = 69 We can rearrange this to solve for the equilibrium constant, K: K = pHI2 / (pH2S,pI2) At equilibrium, let's say the pressure of HI is x atm. Therefore, the pressure of H2S and I2 is also x atm. Now we can plug these values into the equation above, along with the given value of the equilibrium constant at 340 K: 69 = x2 / (5.00 - x)2 Solving for x gives us x = 1.09 atm. However, we need to find the total pressure, which is the sum of the pressures of all the gases in the system. Therefore, the total pressure at equilibrium is: 2x + x + x = 4.09 atm ≈ 2.18 atm (rounded to two significant figures)

At 340 K, Kp = 69 for the Reaction H2(g)

At 340 K, K_p = 69 for the Reaction H₂(g)