Dimethyl ether, CH3OCH3, is used as an environmentally friendly propellant in spray cans - VCE - SSCE Chemistry - Question 3 - 2009 - Paper 1

For the reverse reaction:

$CH_3OCH_3(g) + H_2O(g) \leftrightarrow 2CH_3OH(g)$

the equilibrium constant can be calculated using the inverse of Kc:

$K' = \frac{1}{K_c} = \frac{1}{5.74} = 0.174$

i. Calculate the concentration, in mol L⁻¹, of methanol at equilibrium.

To find the concentration, use the formula:

$C = \frac{n}{V}$

where:

• n = amount of substance in mol = 0.340 mol
• V = volume in L = 20.0 L

Thus,

$C = \frac{0.340 \text{ mol}}{20.0 \text{ L}} = 0.0170 \, \text{mol L}^{-1}$

ii. Calculate the amount, in mol, of dimethyl ether present at equilibrium.

Given the stoichiometry from reaction:

For every 1 mol of dimethyl ether produced, 2 mol of methanol is consumed. Therefore:

Let x be the concentration of dimethyl ether at equilibrium:

At equilibrium: $[CH_3OH] = 0.340 - 2x$ Assuming all methanol is converted into dimethyl ether, the amount of dimethyl ether produced = x mol. Thus,

At equilibrium, $[CH_3OH] = 0.340 - 2(0.0170) = 0.3060 \, \text{mol}$

So, $x = 0.0170 \, \text{mol}$

iii. Calculate the amount, in mol, of methanol initially pumped into the reaction vessel. Initially, let's denote the initial moles of methanol as n-0:

Let n-0 = n + x:

Where n = moles of methanol remaining, x = moles of dimethyl ether obtained,

i.e. $n_0 = 0.340 + 0.0170 = 0.357 \, \text{mol}$