The complete question is as follows: At 700 K,
decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of
at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of
is 0.79 atm.
Explanation:
The equation for decomposition of
is as follows.
![CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)](https://tex.z-dn.net/?f=CCl_%7B4%7D%28g%29%20%5Crightleftharpoons%20C%28s%29%20%2B%202Cl_%7B2%7D%28g%29)
Let us assume that initial concentration of
is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
![CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)](https://tex.z-dn.net/?f=CCl_%7B4%7D%28g%29%20%5Crightleftharpoons%20C%28s%29%20%2B%202Cl_%7B2%7D%28g%29)
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](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7BP%5E%7B2%7D_%7BCl_%7B2%7D%7D%7D%7BP_%7BCCl_%7B4%7D%7D%7D%5C%5C0.76%20%3D%20%5Cfrac%7B%282x%29%5E%7B2%7D%7D%7B%28a%20-%20x%29%7D%5C%5C0.76%20%3D%20%5Cfrac%7B4x%5E%7B2%7D%7D%7B1.1%20-%20x%20-%20x%7D%5C%5C0.76%20%3D%20%5Cfrac%7B4x%5E%7B2%7D%7D%7B1.1%20-%202x%7D%5C%5Cx%20%3D%200.31%20atm)
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
is 0.79 atm.