QUESTION 121 (5 marks)
A closed system is at equilibrium for . A catalyst is added without changing temperature, volume or composition. The rates just before and just after addition are shown.
| Time | Forward rate / mmol L⁻¹ min⁻¹ | Reverse rate / mmol L⁻¹ min⁻¹ |
|---|---|---|
| Before addition | 2.0 | 2.0 |
| Immediately after addition | 6.0 | 6.0 |
(a) State whether the system has a net forward reaction immediately after the catalyst is added. Justify your answer. [2 marks]
(b) Explain the rate changes in terms of activation energy. [2 marks]
(c) State the effect on the equilibrium concentrations and on Kc. [1 marks]
Practice marking scheme
Answer
There is no net forward reaction: both rates remain equal. A catalyst provides a pathway with lower activation barriers in both directions. Equilibrium concentrations and Kc are unchanged.
Working
Before and after addition, the rate of formation of Y equals its rate of consumption. The catalyst increases both rates by providing an alternative pathway with lower activation energies. At the same temperature, a larger fraction of collisions can react. It does not alter the equilibrium composition or equilibrium constant.
Marking criteria
- States that there is no net forward reaction. [1 mark]
- Justifies this using equality of the two rates after addition. [1 mark]
- Identifies an alternative pathway with lower activation barriers in both directions. [1 mark]
- Links lower barriers to more successful collisions at the same temperature. [1 mark]
- States that equilibrium concentrations and Kc remain unchanged. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.