Chapter 6: Problem 41
For a reversible reaction: \(\mathrm{A} \underset{K_{2}}{\stackrel{K_{1}}{K}} \mathrm{~B}\), the initial molar concentration of \(\mathrm{A}\) and \(\mathrm{B}\) are \(a \mathrm{M}\) and \(b \mathrm{M}\), respectively. If \(x \mathrm{M}\) of \(\mathrm{A}\) is reacted till the achievement of equilibrium, then \(x\) is (a) \(\frac{K_{1} a-K_{2} b}{K_{1}+K_{2}}\) (b) \(\frac{K_{1} a-K_{2} b}{K_{1}-K_{2}}\) (c) \(\frac{K_{1} a-K_{2} b}{K_{1} K_{2}}\) (d) \(\frac{K_{1} a+K_{2} b}{K_{1}+K_{2}}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.