Respuesta :
Answer: The value of [tex]K_p[/tex] for the given equation is 3.065
Explanation:
To calculate the partial pressure of phosgene we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = ?
V = Volume of the gas = 1.50 L
T = Temperature of the gas = 700 K
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
n = number of moles of phosgene = 0.0340 moles
Putting values in above equation, we get:
[tex]p_{COCl_2}\times 1.50L=0.0340\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 700K\\\\p_{COCl_2}=\frac{0.0340\times 0.0821\times 700}{1.50}=1.303atm[/tex]
As, initially phosgene is present in the system. So, the reaction is going backwards.
We are given:
Equilibrium partial pressure of CO = 0.509 atm
For the given chemical equation:
[tex]CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)[/tex]
Initial: 1.303
At eqllm: x x 1.303-x
Evaluating the value of 'x'
[tex]\Rightarrow x=0.509[/tex]
So, equilibrium partial pressure of chlorine gas = x = 0.509 atm
Equilibrium partial pressure of phosgene = 1.303 - x = [1.303 - 0.509] = 0.794 atm
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{0.794}{0.509\times 0.509}\\\\K_p=3.065[/tex]
Hence, the value of [tex]K_p[/tex] for the given equation is 3.065
