Question

When heated, iodine vapor dissociates according to this equation: I_{2 (g)} ⇌ 2I (g). At 1274K a sample exhibits a partial pressure of I₂ of 0.1122 and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, K_p for the decomposition at 1274K

Short answer

The value of equilibrium constant K_p=0.17.

Step-by-step solution

Define the Equilibrium constant

The Definition of equilibrium constant: a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature

Calculation of equilibrium constant

The need to calculate the equilibrium constant K_p for this reaction:

I_{2 (g)}⇌ 2I (g).

We calculate the value K_p

\({K_p} = \frac{{p(I)_{eq}^2}}{{p{{\left( {{I_2}} \right)}_{eq}}}}\)

The equilibrium constant of the compound

The p_eq is equilibrium pressure of the compound. We are given all the data in the task:

\begin{aligned}{p{{(I)}_{eq}}=0.1378{\rm{atm}}}\\{p{{\left({{I_2}}\right)}_{eq}}=0.1122{\rm{atm}}}\end{aligned}

We calculate K_p

\begin{aligned}{{K_p}=\frac{{{{0.1378}^2}}}{{0.1122}}}\\{{K_p}=0.17}\end{aligned}