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Chapter 13: Q58 E (page 717)

When heated, iodine vapor dissociates according to this equation: I2 (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, Kp for the decomposition at 1274K

Short Answer

Expert verified

The value of equilibrium constant Kp=0.17.

Step by step solution

01

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

02

Calculation of equilibrium constant

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

I2 (g)⇌ 2I (g).

We calculate the value Kp

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

03

The equilibrium constant of the compound

The peq 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 Kp

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

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