$$\ce{CH3OH(g) + NOCl(g) <=> CH3ONO(g) + HCl(g)}$$ The volume of the container is $\pu{433 cm^3},$ $T = \pu{50 °C}.$ Methanol was added until the pressure was $\pu{50.1 mbar},$ then $\pu{0.059 g}$ of $\ce{NOCl}$ was added. At equilibrium the partial pressure of $\ce{NOCl}$ is $\pu{27.6 mbar}.$ Calculate $K.$

I assumed that $\pu{50.1 mbar}$ was the partial pressure of methanol (I do not know if this is correct) and calculated the initial partial pressure for $\ce{NOCl}$ from the ideal gas law:

$$p = \frac{nRT}{V} = \frac {(\pu{0.059 g}/\pu{65 g/mol})\times 8.314\times 323}{\pu{4.33E-1}} = \pu{5.63 Pa} = \pu{5.63E-5 bar}$$


$$ \begin{array}{cccc} &\ce{CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)} \\ &\pu{0.0501 bar} && \pu{5.63E-3 bar} && 0 && 0 \\ &\pu{0.0501 bar} - x && \pu{0.0276 bar} && x && x\\ \end{array} $$

When I then solve for $x$ I get a negative value which isn't correct. Is $\pu{50.1 mbar}$ supposed to be the total pressure and if yes, how do I use that then? The correct answer for $K$ is $1.33.$

  • 1
    $\begingroup$ Upon looking at your table I immediately noticed that you initially start out with 0.00563 bar of NOCl and end up with 0.0276. How is possible to end up with more than you started with? $\endgroup$
    – MaxW
    Commented Jan 29, 2021 at 13:23
  • 1
    $\begingroup$ Your calculation is wrong. You report a volume of $0.433 L$. Here the volume should be in cubic meter. So the volume should be $\ce{4.33·10^{-4} m^3}$ and the initial pressure of $\ce{NOCl}$ is $5630 Pa = 0.056 bar $. With this value, you can obtain a reasonable value for $x$.. $\endgroup$
    – Maurice
    Commented Jan 29, 2021 at 13:31

1 Answer 1


As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$:

$$ \begin{array}{lccccc} &\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\ &\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) && 0 && 0 \\ &\text{C} && -x && -x && x && x \\ &\text{E} && p_0(\ce{CH3OH}) - x && p_0(\ce{NOCl}) - x && x && x \end{array} $$

You are interpreting given conditions wrong. The problem explicitly provides you with the initial partial pressure of methanol $p_0(\ce{CH3OH}) = \pu{50.1 mbar}.$ As for the initial partial pressure of nitrosyl chloride, it can be found applying the ideal gas law:

$$ \begin{align} p_0(\ce{NOCl}) &= \frac{m(\ce{NOCl})}{M(\ce{NOCl})}\frac{RT}{V}\\ &= \frac{\pu{0.059 g}}{\pu{65.46 g mol^-1}}\frac{\pu{8.314 J mol^-1 K^-1}\times\pu{323 K}}{\pu{4.33E-4 m^3}} \\ &\approx \pu{5590 Pa} \\ &\approx \pu{55.9 mbar} \tag{1} \end{align} $$

Note that methanol and nitrosyl chloride reaction mix is apparently not equimolar. Also, the total pressure is not relevant for obtaining the solution. Do keep in mind though it is not the initial partial pressure of the first component because the second gaseous reagent has been added afterwards.

The unknown $x$ can be found trivially since you are given the equilibrium partial pressure of nitrosyl chloride $p(\ce{NOCl}) = \pu{27.6 mbar}$:

$$ \begin{align} p_0(\ce{NOCl}) - x = p(\ce{NOCl}) \quad\implies\quad x &= p_0(\ce{NOCl}) - p(\ce{NOCl}) \\ &= \pu{55.9 mbar} - \pu{27.6 mbar} \\ &= \pu{28.3 mbar} \tag{2} \end{align} $$

Finally we can write an expression for $K_p$ (by the way, here $K_p = K_c$ since $\Delta n = 0)$:

$$ \begin{align} K_p &= \frac{p(\ce{CH3ONO})\cdot p(\ce{HCl})}{p(\ce{CH3OH})\cdot p(\ce{NOCl})}\\ &= \frac{x^2}{(p_0(\ce{CH3OH}) - x)\cdot p(\ce{NOCl})}\\ &= \frac{(\pu{28.3 mbar})^2}{(\pu{50.1 mbar} - \pu{28.3 mbar})\times\pu{27.6 mbar}} \\ &\approx 1.33 \tag{3} \end{align} $$

Don't forget to do dimensional analysis to check yourself and never omit units in your calculation. Also, keep an eye on significant figures: the number for the molar mass you used was too sloppy; instead of two figures $(\pu{65 g mol^-1})$ you had to go with four $(\pu{65.46 g mol^-1}).$


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