# Calculate the heat change (ΔΗ°rxn) for the slow reaction of zinc with water?

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Zn(s)+2H2​O(l)–> Zn2+(aq)+2OH−(aq)+H2​(g)

Using data from the following reactions and applying Hess’s law,

H+(aq)+OH−(aq) –> H2​O(l)

ΔH^o rxn1​ = −56.0 kJ

Zn(s) –> Zn2+(aq)

ΔH ^o rxn2​ = −153.9 kJ

1/2 ​H2​(g) –> H+(aq)

ΔH ^o rxn3 = 0.0kJ

1. ΔH ^o rxn​ = ??? kJ

• To use hess’ law you need to get all the reactants on one side and products and the other side by adding the reactants you have below. Whenever you flip and equation around you need to change the sign of the enthalpy of the reaction and whenever you multiply or divide the reaction by a number you need to do the same to the enthalpy of the reaction.

For this problem:

you want Zn(s) on the left side so leave the second equation as it is.

Zn(s) –> Zn2+(aq) ΔΗ°rxn = -153.9 kJ

you want two H2O (l) on the left side so you have to flip the first equation and multiply it by two.

2H2O (l) –> 2OH-(aq) + 2H+(aq) ΔΗ°rxn = +56*2 kJ

finally you want two H2(g) on the right so you multiply the third equation by two

2H+ (aq) –> 2H2(g) ΔΗ°rxn = 0.0 * 2 kJ

Last step is you have all the equations the way you want them so add them together. Anything that is on the left and right side will cancel (the 2H+(aq) cancels) and then you add the ΔΗ°rxn’s up to get the final answer (-153.9 kJ + 56(2) kJ + 0.2(2) kJ = -41.9 kJ

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