# Question #8f2ab

Apr 30, 2017

The mixture is at equilibrium, so a change in volume/pressure will cause the total moles of reactants and products to change.

#### Explanation:

We can find the new pressure of the system after the volume is reduced by using the equation:

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

$\left(1.2 a t m\right) \left(2.0 L\right) = \left({P}_{2}\right) \left(1.0 L\right)$

${P}_{2} = \frac{\left(1.2 a t m\right) \left(2.0 L\right)}{1.0 L}$

${P}_{2} = 2.4 a t m$

But since CO(g) and H2(g) react to form an equilibrium, given by the formula:

$C O \left(g\right) + 2 {H}_{2} \left(g\right) \rightarrow C {H}_{3} O H \left(g\right)$

The increased pressure will cause more of the reactants to form the product, therefore reducing the pressure (because 3 mols reactant will form 1 mol product).

So the final pressure will be more than 1.2 atm but less than the predicted 2.4 atm.