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Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your answer numerically using two significant figures. hints

Carlton Burgess

in Chemistry

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Justin Parker on April 16, 2018

We cannot solve this problem without the use of empirical data. These reactions have already been experienced by the scientists. The standard Gibb's free energy, ΔG°, (which occur at the standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.the glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=-7.28 kJ/mol fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=-1.67 kJ/molTherefore, the reaction is a two-step process in which the glucose-6-phosphate is the intermediate product.the glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate In this case, you just add the ΔG°. However, given that we have the reverse of the second reaction, to end with the terminal product, fructose-6-phosphate, you will have to take the opposite sign of ΔG°.ΔG°,total = -7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/molThen, the equation to relate ΔG° to the equilibrium constant K isΔG° = -RTlnK, where R is the gas constant, equal to 0.008317 kJ/mol K.-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)lnK = 2.2635 K = e^2.2635 K = 9.62


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