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.

Step-by-step 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 = ((2x)^(2))/((a - x))\\0.76 = (4x^(2))/(1.1 - x - x)\\0.76 = (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.