Question

Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.

Answer 1

The complete question is as follows: At 700 K, $CCl_{4}$ decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of $CCl_{4}$ at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of $CCl_{4}$ is 0.79 atm.

Explanation:

The equation for decomposition of $CCl_{4}$ is as follows.

$CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)$

Let us assume that initial concentration of $CCl_{4}$ is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   $CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)$

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

$K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm$

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of $CCl_{4}$ is 0.79 atm.