Reversibility and dynamic equilibria
3 marks · NormalAt dynamic equilibrium in a sealed vessel, the following reversible reaction occurs:
Suppose the rate of the forward reaction is 0.160 mol dm⁻³ min⁻¹ when equilibrium is reached. Calculate:
The rate of the reverse reaction at equilibrium.
If 0.160 mol of N2O4 is used up in 1 minute, how much NO2 would be produced in that time by the forward reaction?
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 0.160 mol dm⁻³ min⁻¹
(b) 0.320 mol NO2 mol
Common mistakes: Dividing rather than multiplying stoichiometric ratio; Mismatching rates for forward and reverse
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States rate of reverse reaction is equal to forward at equilibrium: 0.16 mol dm⁻³ min⁻¹
M1 (b) · 1 mark · Correctly uses stoichiometry: 1 mol N2O4 → 2 mol NO2
A1 (b) · 1 mark · Final answer: 0.32 mol NO2 formed
Worked solution
a) At equilibrium, the forward and reverse reaction rates are equal. So reverse rate = 0.160 mol dm⁻³ min⁻¹.
b) From the equation, 1 mol N2O4 forms 2 mol NO2. If 0.160 mol N2O4 reacts, it produces: 0.160 × 2 = 0.320 mol NO2.