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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.

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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.

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