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Given that $K=18$ for the equilibrium below, calculate the total pressure in a sealed vessel initially containing only $\ce{NH4I(s)}$, that reaches equilibrium.

$$\ce{NH4I(s) \leftrightharpoons NH3(g) + HI(g)}$$

a) $4.2\mathrm{~atm}$
b) $8.5~\mathrm{atm}$
c) $2.1~\mathrm{atm}$
d) $\mathrm{18~atm}$
e) $\mathrm{6.3~atm}$


Using $K = \ce{[NH3][HI]} = 18$, I found $\ce{[NH3]=[HI]=}~\frac{4.24mol}{L}$, and therefore $8.48~\mathrm{mol}$ of gas is produced, but I don't know how to find the equilibrium pressure without temperature and volume.

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I think what is given here is $K_p= p(\ce{NH3}). p(\ce{HI})$ and not $K_c$. So, What you have here are the partial pressures of the gases. As you have a sealed vessel, so the sum of the partial pressures of the gases gives the total pressure, 8.5 atm

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