Reversible Reactions and Equilibria
7 marks · HardConsider the industrial reaction below, carried out in a sealed vessel:
This reaction can reach dynamic equilibrium. The mixture initially contains only X.
Describe what happens to the concentration of X as the system approaches equilibrium. Explain your answer.
At equilibrium, more X is added. Predict what will happen to the concentrations of Y and Z and explain your answer in terms of the position of equilibrium.
Suggest one reason why maintaining constant concentrations of products in continuous industrial reactions is important.
Explain, using the idea of equilibrium, what would happen to the rate of the forward and reverse reactions immediately after X is added, until the new equilibrium is reached.
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.
Show answer
(a) The concentration of X decreases as it reacts to form Y and Z, until the rate of the forward reaction equals the backward reaction. When equilibrium is reached, the concentration of X remains constant. · must show Decrease, then constant at equilibrium, both explained.
(b) When more X is added, the equilibrium shifts right, so the concentrations of Y and Z increase as extra X is converted into Y and Z to oppose the change. · must show Shift prediction and explanation.
(c) Constant product concentrations are needed for consistent product supply and efficient process control in industry. · must show Mention of consistency or efficiency.
(d) Immediately after adding X, the forward rate increases as X is high, but the reverse rate is unchanged. As more Y and Z form, the reverse rate rises, and both rates eventually become equal again at the new equilibrium. · must show Initial increase in forward rate, unchanged reverse, then both equal at new equilibrium.
Also accept: Clear, chemically valid alternatives accepted.
Common mistakes: Assuming concentrations must be equal at equilibrium.; Not linking rate changes to concentration changes.
Show mark scheme and worked solution
Marking points
C1 (a) · 1 mark · States that X decreases as the forward reaction occurs initially.
C2 (a) · 1 mark · Explains equilibrium—X remains constant when rates are equal.
C3 (b) · 1 mark · Predicts that adding X shifts equilibrium right and Y and Z increase.
C4 (b) · 1 mark · Explains reason in terms of position of equilibrium opposing change.
C5 (c) · 1 mark · Gives a valid industrial reason (e.g. consistency, efficiency).
C6 (d) · 1 mark · States forward rate increases, reverse is unchanged immediately after X is added.
C7 (d) · 1 mark · Explains that both rates eventually become equal at new equilibrium.
Worked solution
a) X decreases initially as it reacts to form Y and Z; once equilibrium is reached, its concentration remains constant because the forward and reverse reaction rates are equal.
b) Adding X increases its concentration; equilibrium shifts right, making more Y and Z to oppose the change, so their concentrations rise.
c) This allows the process to deliver a steady supply of product and maintain process control, which is important for industrial efficiency.
d) Immediately after X is added, the forward reaction rate increases because X concentration is higher, while the reverse rate stays the same. As Y and Z build up, the reverse rate also increases, and eventually both reach a new equal rate at the new equilibrium.