The reversible reaction 2H 2 (g) + CO(g) <-> CH 3 OH(g) + heat is carried out by

Question

The reversible reaction 2H2 (g) + CO(g) <-> CH3OH(g) + heat is carried out by mixing carbon monoxide and hydrogen gases in a closed vessel under high pressure with a suitable catalyst. After equilibrium is established at high temperature and pressure, all three substance are present. If the pressure on the system is lowered, with the temperature kept constant, what will be the result?

(a) The amount of CH3OH will be increased
(b) The amount of CH3OH will be decreased
(c) The amount of each substance will be unchanged
(d) The amount of each substance will be increased

The answer doesn't make sense. Based on gas-phase equilibria, if the volume is decreased, the reaction will shift towards the side of the reaction that has fewer gaseous particles. The product side has fewer gaseous particles! So the amount of CH3OH should increase. Please explain!

Thanks!

Explanation / Answer

Given, 2H2 (g) + CO(g) <-> CH3OH(g)

reactants : 2 mole of H2 and 1 mole of CO = 3 moles total

products : 1 mole of CH3OH

From, Boyle's Law, at constant Temperature and number of moles of gas, Pressure and volume are inversely related.

so, PV = const. => P 1/V

Here, Pressure is reduced so, Volume will increase.(But not decrease)

Therefore, According to Le Chatelier's Principle

The reaction will move towards more moles of gas molecules as pressure of the system decrease or volume of the system increases.

Therefore, The amount of CH3OH will be decreased as it is less in number of moles.

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