Question #8f2ab

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Question #8f2ab

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Answer 1 · 2017-04-30T19:44:44

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:

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1 = P_2V_2$

$(1.2 a t m) (2.0 L) = (P_{2}) (1.0 L)$ $(1.2atm)(2.0L) = (P_2)(1.0L)$

$P_{2} = \frac{(1.2 a t m) (2.0 L)}{1.0 L}$ $P_2 = ((1.2atm)(2.0L))/(1.0L)$

$P_{2} = 2.4 a t m$ $P_2 = 2.4 atm$

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

$C O (g) + 2 H_{2} (g) \to C H_{3} O H (g)$ $CO(g) + 2H_2(g) rarr CH_3OH(g)$

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.

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Question #8f2ab

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