Question

Consider the reaction below: i₂(g) ⇌ 2 i(g). The Kc of this reaction is 3.820 × 10⁻⁵ at 791.0 °C. Calculate Kp at the same temperature for the reverse reaction (r = 0.08314 L⋅bar/mol⋅K). 2 i(g) ⇌ i₂(g).

Answer 1

Kp for the reverse reaction 2 I(g) ⇌ I_2(g) is calculated using the provided Kc value for the forward reaction, the ideal gas constant, and the temperature in Kelvin, taking into account the change in moles of gas (Δn).

Step-by-step explanation:

The question pertains to calculating the equilibrium constant (Kp) for the reverse reaction of iodine dissociation into iodine atoms at a given temperature, where the equilibrium constant (Kc) for the forward reaction is provided. Given the reaction I2(g) ⇌ 2 I(g) with Kc = 3.820 × 10−5 at 791.0 °C, we are asked to calculate Kp for the reverse reaction using the equation Kp = Kc(RT)Δn, where Δn is the change in moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

To find Kp for the reverse reaction, first note that Δn = -1 (since the reaction goes from 2 moles of gaseous iodine atoms to 1 mole of gaseous iodine molecules). Using the given value for the gas constant (R = 0.08314 L·bar/mol·K), and converting the temperature to Kelvin (T = 791.0 + 273.15 = 1064.15 K), we apply the formula:

Kp = Kc(RT)Δn = 3.820 × 10−5(0.08314 * 1064.15)−1 = (3.820 × 10−5 / (0.08314 * 1064.15)) = Kp for the reverse reaction.