Methane, CH4, reacts with I2 according to the reaction CH4(g) + I...

Question

Methane, CH4, reacts with I2 according to the reaction CH4(g) + I2(g) ⇌ CH3I(g) + HI(g). At 630 K, Kp for this reaction is 2.26 × 10^-4. A reaction was set up at 630 K with initial partial pressures of methane at 105.1 torr and I2 at 7.96 torr. Calculate the pressures, in torr, of all reactants and products at equilibrium.

Accepted Answer

Step 1: Write the expression for the equilibrium constant Kp for the reaction. For the reaction CH4(g) + I2(g) ⇌ CH3I(g) + HI(g), the expression is Kp = (P_CH3I * P_HI) / (P_CH4 * P_I2).. Step 2: Define the change in pressure for each species in terms of a variable, x. Assume that x is the change in pressure for CH4 and I2 as they react to form CH3I and HI. Therefore, the changes are: CH4: 105.1 - x, I2: 7.96 - x, CH3I: x, HI: x.. Step 3: Substitute the equilibrium pressures in terms of x into the Kp expression. This gives Kp = (x * x) / ((105.1 - x) * (7.96 - x)).. Step 4: Set the Kp expression equal to the given Kp value, 2.26 × 10^-4, and solve the resulting quadratic equation for x. This will involve expanding the equation and using the quadratic formula.. Step 5: Calculate the equilibrium pressures of all species using the value of x obtained from the quadratic equation. The pressures will be: P_CH4 = 105.1 - x, P_I2 = 7.96 - x, P_CH3I = x, P_HI = x.