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Ch.15 - Chemical Equilibrium
All textbooksBrown 14th EditionCh.15 - Chemical EquilibriumProblem 79
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Chapter 15, Problem 79

For the equilibrium 2 IBr(g) ⇌ I2(g) + Br2(g), Kp = 8.5 * 10^-3 at 150 _x001F_C. If 0.025 atm of IBr is placed in a 2.0-L container, what is the partial pressure of all substances after equilibrium is reached?

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1
Step 1: Write the expression for the equilibrium constant Kp for the reaction 2 IBr(g) ⇌ I2(g) + Br2(g). The expression is Kp = (P_{I2} * P_{Br2}) / (P_{IBr}^2), where P represents the partial pressures of the gases.
Step 2: Define the initial conditions. Initially, the partial pressure of IBr is 0.025 atm, and the partial pressures of I2 and Br2 are 0 atm since they are not present initially.
Step 3: Set up an ICE (Initial, Change, Equilibrium) table to track the changes in partial pressures. Let x be the change in pressure for I2 and Br2, and 2x for IBr since it dissociates into two moles of IBr for every mole of I2 and Br2 formed.
Step 4: Express the equilibrium partial pressures in terms of x. At equilibrium, P_{IBr} = 0.025 - 2x, P_{I2} = x, and P_{Br2} = x.
Step 5: Substitute the equilibrium partial pressures into the Kp expression and solve for x. Use the equation Kp = (x * x) / (0.025 - 2x)^2 = 8.5 * 10^-3 to find the value of x, which will allow you to calculate the equilibrium partial pressures of IBr, I2, and Br2.
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