Question

Calculate the heat change (ΔH°rxn) for the slow reaction of zinc with water Zn(s) + 2H2O(l) -> Zn2+(aq) + 2OH-(aq) + H2(g) Using data from the following reactions and applying Hess's law, H+(aq) + OH-(aq) -> H2O(l) ΔΗ°rxn1 = -56.0 kJ Zn(s) -> Zn2+(aq) ΔΗ°rxn2 = -153.9.0 kJ

1/2 H2(g) -> H+(aq) ΔΗ°rxn3 = 0.0 kJ

Answer 1

To calculate the heat change (ΔH°rxn) for the reaction of zinc with water, we can use Hess's law and the given reactions. The enthalpy change for the reaction between H+(aq) and OH-(aq) is -56.0 kJ (ΔH°rxn1), the enthalpy change for the reaction between Zn(s) and Zn2+(aq) is -153.9 kJ (ΔH°rxn2), and the enthalpy change for the reaction between 1/2 H2(g) and H+(aq) is 0.0 kJ (ΔH°rxn3). By summing the enthalpy changes of the individual reactions, we can calculate the enthalpy change for the overall reaction, which is -209.9 kJ.

Step-by-step explanation:

To calculate the heat change (ΔH°rxn) for the reaction of zinc with water, we can use Hess's law and the given reactions. The enthalpy change for the reaction between H+(aq) and OH-(aq) is -56.0 kJ (ΔH°rxn1), the enthalpy change for the reaction between Zn(s) and Zn2+(aq) is -153.9 kJ (ΔH°rxn2), and the enthalpy change for the reaction between 1/2 H2(g) and H+(aq) is 0.0 kJ (ΔH°rxn3).

Using these values, we can rearrange and sum the reactions to get the desired reaction:

Zn(s) + 2H2O(l) -> Zn2+(aq) + 2OH-(aq) + H2(g)

By adding the enthalpy changes of the individual reactions, we can calculate the enthalpy change for the overall reaction:

-153.9 kJ + (-56.0 kJ) + (0.0 kJ) = -209.9 kJ

Answer 2

The heat change (ΔH°rxn) for the reaction of zinc with water, calculated using Hess's law and given reaction data, is -209.9 kJ.

Step-by-step explanation:

To calculate the heat change (ΔH°rxn) for the reaction of zinc with water using Hess's law, we need to manipulate the given equations to arrive at the target equation. The reaction of interest is:
Zn(s) + 2H2O(l) -> Zn2+(aq) + 2OH-(aq) + H2(g).

We have the following known reactions and their ΔH° values:
1. H+(aq) + OH-(aq) -> H2O(l) ΔH°rxn1 = -56.0 kJ
2. Zn(s) -> Zn2+(aq) ΔH°rxn2 = -153.9 kJ
3. 1/2 H2(g) -> H+(aq) ΔH°rxn3 = 0.0 kJ.

By Hess's law, summing the ΔH° values of reactions 1 and 2 while adding twice the ΔH° value of reaction 3, because the balanced equation has 1 mole of H2 on the product side, gives us the overall ΔH°rxn for the target equation. Thus, the heat change is obtained as follows:

ΔH°rxn = ΔH°rxn1 + ΔH°rxn2 + 2(ΔH°rxn3)
ΔH°rxn = (-56.0 kJ) + (-153.9 kJ) + 2(0.0 kJ)
ΔH°rxn = -209.9 kJ

The calculated heat change for the reaction of zinc with water is -209.9 kJ.