Chapter 1: Q14P (page 21)

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

Short Answer

Expert verified

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

Step by step solution

Given data

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

$\begin{array}{l} Δ G^{0} = - 20.9 k g / m o l \\ T = 25 ° C = 298 K \\ R = 8.31 J / m o l K \end{array}$

Formula and calculation for equilibrium constant

$Δ G^{0} = - R T \ln K$

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

Conclusion

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

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

Short Answer

Step by step solution

Given data

Formula and calculation for equilibrium constant

Conclusion

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