Chapter 7, Problem 7.76P

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction should be calculated.

Concept Introduction:

The given equilibrium reaction representing the conversion of glucose-1-phosphate to glucose-6-phosphate,

$\text{Glucose}-1-\text{phosphate}◰\hspace{1em}\text{Glucose}-6-\text{phosphate}$

Equilibrium constant, K for this reaction can be determined using the equilibrium molar concentrations of glucose-1-phosphate and glucose-6-phosphate as below,

$\text{K=}\frac{\text{[Glucose-6-phosphate]}}{\text{[Glucose-1-phosphate]}}$.

Answer to Problem 7.76P

The equilibrium constant for the given equilibrium is 24.

Explanation of Solution

The given equilibrium reaction representing the conversion of glucose-1-phosphate to glucose-6-phosphate:

$\text{Glucose}-1-\text{phosphate}◰\hspace{1em}\text{Glucose}-6-\text{phosphate}$

Initial concentration of glucose-1-phosphate is 0.10 M.

Equilibrium concentration of glucose-6-phosphate is $9.6\times {10}^{-2}M$

Setting up an ICE table for the equilibrium,

$\begin{array}{l}\text{Glucose}-1-\text{phosphate}◰\hspace{1em}\text{Glucose}-6-\text{phosphate}\\ \text{Initial(M)}\text{0}\text{.10}\text{0}\\ \text{Change(M)}-\text{X}+\text{X}\\ \text{Equilibrium(M})(0.10-\text{X})\text{X}\end{array}$

$\text{K=}\frac{\text{X}}{(0.10-\text{X})}$

$\begin{array}{l}\text{[Glucose-6-phosphate] = }\text{X}=9.6\times {10}^{-2}M\\ \text{[Glucose-1-phosphate] = (0}\text{.10}-\text{X}\text{)}\text{=}(\text{0}\text{.10}-9.6\times {10}^{-2})=0.004M\end{array}$

$\text{K=}\frac{(9.6\times {10}^{-2})}{(0.004)}$

  = 24.

Conclusion

Therefore, the equilibrium constant for the given equilibrium will be 24.