2SO2(g) + O2(g) ↔ 2SO3(g) ΔH = -189 kJ mol-1 The equilibrium constant for the reaction above is increased by

Assessment: JAMB UTME - Chemistry - 2001 Subject: Chemistry

Question 1 Report

2SO2(g) + O2(g) ↔ 2SO3(g)

ΔH = -189 kJ mol-1

The equilibrium constant for the reaction above is increased by

Answer Details
The equilibrium constant for the reaction given is a measure of the ratio of the concentration of the products to the reactants at equilibrium, and it is determined by the temperature and pressure of the system. The value of the equilibrium constant depends on the enthalpy change of the reaction, which is given as -189 kJ/mol in this case. According to Le Chatelier's principle, the equilibrium constant can be increased by shifting the equilibrium towards the products. This can be achieved by decreasing the temperature or increasing the pressure of the system. The addition of a catalyst to the system can increase the rate of the reaction but it will not affect the equilibrium constant. Increasing the surface area of the vessel will not have any significant effect on the equilibrium constant either. Therefore, the correct answer is "Increasing the pressure of the system."

Download The App On Google Playstore

Everything you need to excel in your exams

Green Bridge CBT Mobile App
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning