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Question 1: Given the following reactions and standard molar enthalpy changes ΔrH0: D-glucose 6-phosphate2-(aq) = D-fructose...

Question 1: Given the following reactions and standard molar enthalpy changes ΔrH0:

D-glucose 6-phosphate2-(aq) = D-fructose 6-phosphate2-(aq) (1)

D-glucose + ATP4-(aq) = D-glucose 6-phosphate2-(aq) + ADP3-(aq) + H+(aq) (2)

D-fructose + ATP4-(aq) = D-fructose 6-phosphate2-(aq) + ADP3-(aq) + H+(aq) (3)

ΔrH0 (1) = 11.7 kJ mol-1

ΔrH0 (2) = -23.8 kJ mol-1

ΔrH0 (3) = -15.0 kJ mol-1

Calculate ΔrH0 for the reaction: D-glucose(aq) = D-fructose(aq) (4)

Expert Solution

Solution :-

In the final reaction we want D-glucose on the reactant side and D- fructose on the product side

Therefore we need to rearrange the other three equations in a such a way that after adding those equation we end with the final desired equation

So we need to reverse the equation 3

And keep the equation 1 and 2 as it is then we get new equations

When the equation is reversed then value of the delta H rxn remains the same but its sign changes and while adding the equations we add up the delta Hrxn values to get the enthalpy change for the final equation .

Following image shows the rearrangement of the equations

Therefore the Delta Ho rxn for the final equation is 2.9 kJ/mol

queen_honey_blossom answered 2 years ago

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