Question

Calculate the equilibrium constant for the reaction. Glucose-1-phosphate+H₂O gives Glucose+ H₂PO₄^- at pH 7 and 25^oC.

Short answer

Glucose-1-phosphate on hydrolysis yields Glucose and Phosphate at pH 7 is $\mathrm{K}=4.627\times {10}^3$.

Step-by-step solution

Given data

Glucose-1-phosphate on hydrolysis yields Glucose and Phosphate.

$\begin{array}{l}\Delta G^0=-20.9kg/mol\\ T=25°C=298K\\ R=8.31J/molK\end{array}$

Formula and calculation for equilibrium constant

$\Delta G^0=-RT\ln K$

$\begin{array}{l}-20.9\times {10}^3=-8.31\times 298\ln K\\ \ln K=8.439\\ K=e^{8.439}=4.627\times {10}^3\end{array}$

Conclusion

Therefore, the equilibrium constant for the reaction Glucose-1-phosphate on hydrolysis yields Glucose and Phosphate at pH 7 is $K=4.627\times {10}^3$.