I have the reaction $\ce{CO_{(g)} + Cl2_{(g)} <=> COCl2_{(g)}}$ where $K=4.5\times10^9$ at $100\ \mathrm{^\circ C}$

The first question was to find $K_p$ which I know is $K_p=K(RT)^{\Delta n}$ where $R=0.08206$, $T=273+100=373$ and $n=1-2=-1$. Put in the numbers and get $K_p\approx1.47\times10^8$.

Now comes the second question and this is where I get stuck:

Equal moles of $\ce{CO}$ and $\ce{Cl2}$ are reacted at $100\ \mathrm{^\circ C}$. If the total pressure at equilibrium is 5.0 atm calculate the equilibrium partial pressures of the gases.

I have tried several solutions, where I either end up with a wrong answer or hit a dead end. I just don't understand how to solve this.


1 Answer 1


The equations you need are $$K_p=\frac{p_{\ce{COCl2}}}{p_{\ce{CO}} p_{\ce{Cl2}}}=1.47\cdot10^8$$ and $$p_\text{tot}=p_{\ce{COCl2}}+p_{\ce{CO}}+p_{\ce{Cl2}}=5.0~\text{atm}$$ Additionally, $n_{\ce{CO}}=n_{\ce{Cl2}}\implies p_{\ce{CO}}=p_{\ce{Cl2}}$, making the system solvable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.