Dimethyl ether, CH3OCH3, is used as an environmentally friendly propellant in spray cans. It can be synthesized from methanol according to the following equation. 2CH3OH(g) ⇌ CH3OCH3(g) + H2O(g); ΔH = -24 kJ mol⁻¹ The equilibrium constant, Kc, for this reaction at 350°C is 5.74. a. Write an expression for K for this reaction. b. Calculate the value of K at 350°C for the following reaction. CH3OCH3(g) + H2O(g) ⇌ 2CH3OH(g) c. Methanol is pumped into an empty 20.0 L reactor vessel. At equilibrium the vessel contains 0.340 mol of methanol at 350°C. i. Calculate the concentration, in mol L⁻¹, of methanol at equilibrium. ii. Calculate the amount, in mol, of dimethyl ether present at equilibrium. iii. Calculate the amount, in mol, of methanol initially pumped into the reaction vessel.

SSCE VCE Chemistry Families of Organic Compounds: Dimethyl ether, CH3OCH3, is used

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

Question 3

Dimethyl ether, CH3OCH3, is used as an environmentally friendly propellant in spray cans. It can be synthesized from methanol according to the following equation. 2... show full transcript

Worked Solution & Example Answer:Dimethyl ether, CH3OCH3, is used as an environmentally friendly propellant in spray cans - VCE - SSCE Chemistry - Question 3 - 2009 - Paper 1

Step 1

a. Write an expression for K for this reaction.

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Answer

The expression for the equilibrium constant, Kc, for the given reaction can be written as:

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

Step 2

b. Calculate the value of K at 350°C for the following reaction.

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Answer

To find K for the reaction:

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

we use the relationship between the equilibrium constants of the forward and reverse reactions:

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

Given that Kc for the forward reaction is 5.74, we have:

$K_{c,reverse} = \frac{1}{5.74} = 0.174$

Step 3

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

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To calculate the concentration of methanol at equilibrium:

$\text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{0.340 \text{ mol}}{20.0 \text{ L}} = 0.0170 \text{ mol L}^{-1}$

Step 4

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

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Using the stoichiometry of the reaction, the initial moles of methanol can be represented as 'X'. At equilibrium:

Let the change in moles of dimethyl ether be 'y'. Therefore:

For methanol:[ 0.340 - 2y ]
For dimethyl ether: [ y ]

Using the equilibrium constant: $K_{c} = \frac{[y][remaining H2O]}{[0.340 - 2y]^2} \to y = 0.144 \text{ mol}$

Step 5

c.iii. Calculate the amount, in mol, of methanol initially pumped into the reaction vessel.

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Answer

To calculate the initial amount of methanol,

Given that 0.340 mol remains in the vessel, and using the stoichiometry of the reaction, we find:

Let 'Z' be the initial amount of methanol, $Z - 2y = 0.340$
Substituting for y gives: $Z = 0.340 + 2(0.144) = 0.628 ext{ mol}$

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

Worked Solution & Example Answer:Dimethyl ether, CH3OCH3, is used as an environmentally friendly propellant in spray cans - VCE - SSCE Chemistry - Question 3 - 2009 - Paper 1

a. Write an expression for K for this reaction.

b. Calculate the value of K at 350°C for the following reaction.

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

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

c.iii. Calculate the amount, in mol, of methanol initially pumped into the reaction vessel.

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