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

The reaction is:

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

Given that the equilibrium constant K for the forward reaction (synthesis of dimethyl ether) is 5.74, we can find K for this reaction using the relationship:

$K_{reverse} = \frac{1}{K_{forward}}$

Thus,

$K = \frac{1}{5.74} \approx 0.174$

To calculate the concentration of methanol at equilibrium:

1. Use the formula for concentration:

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

2. Substitute the number of moles (n = 0.340 mol) and the volume (V = 20.0 L):

   $C = \frac{0.340 \text{ mol}}{20.0 \text{ L}} = 0.017 \text{ mol L}^{-1}$

Using the equilibrium concentrations based on the reaction:

Let x be the amount of dimethyl ether formed:

At equilibrium:

• Concentration of CH3OH = 0.017 mol L⁻¹ (from part c.i)
• The reaction shows that for every 1 mol of dimethyl ether produced, 2 mol of methanol are consumed. Thus:

From the stoichiometry, the concentration of dimethyl ether is:

$[CH_3OCH_3] = \frac{(0.340 - 2x)}{20}$

Given K = 0.174:

$K = \frac{[CH_3OCH_3][H_2O]}{[CH_3OH]^2}$

After solving for x, we can find the total amount in moles.

Assuming there are 0.340 mol at equilibrium, if we denote the initial amount of methanol pumped as n_initial:

From part ii, if we denote the amount of dimethyl ether produced as y:

• Since the equilibrium system might have changed, let's consider:

$n_{initial} = 0.340 + 2y$

We can derive the value of y based on equilibrium constants.

Things will depend on the equilibrium constant from earlier steps to finalize the amount of methanol initially added.