Using data from the following reactions and applying Hess's law, calculate the heat change for the slow reaction of zinc with water (answer must be in kJ/mol Zn):
Zn(s) + 2 H2O(l) ----> Zn^2+(aq) + 2 OH-(aq) + H2(g) (ΔHrxn = ?)
H+(aq) + OH-(aq) ----> H2O(l) (ΔHrxn,1 = -56 kJ/mol H2O)
Zn(s) ----> Zn^2+(aq) (ΔHrxn,2 = -153.9 kJ)
1/2 H2(g) ----> H+(aq) (ΔHrxn,3 = 0.0 kJ)
Using data from the following reactions and applying Hess's law, calculate the heat change for the slow reaction of zinc with water (answer must be in kJ/mol Zn):
Zn(s) + 2 H2O(l) ----> Zn^2+(aq) + 2 OH-(aq) + H2(g) (ΔHrxn = ?)
H+(aq) + OH-(aq) ----> H2O(l) (ΔHrxn,1 = -56 kJ/mol H2O)
Zn(s) ----> Zn^2+(aq) (ΔHrxn,2 = -153.9 kJ)
1/2 H2(g) ----> H+(aq) (ΔHrxn,3 = 0.0 kJ)
To calculate the heat change for the slow reaction of zinc with water, we can apply Hess's law, which states that the enthalpy change for a reaction is independent of the pathway taken.
First, let's determine the overall reaction by combining the given reactions:
Zn(s) + 2 H2O(l) ----> Zn^2+(aq) + 2 OH-(aq) + H2(g)
To simplify the calculations, we can manipulate the reactions and their coefficients to match the overall reaction:
H+(aq) + OH-(aq) ----> H2O(l) (ΔHrxn,1 = -56 kJ/mol H2O)
Zn(s) ----> Zn^2+(aq) (ΔHrxn,2 = -153.9 kJ)
1/2 H2(g) ----> H+(aq) (ΔHrxn,3 = 0.0 kJ)
Now, let's multiply each reaction by the appropriate coefficients to match the coefficients of the overall reaction:
2 * (H+(aq) + OH-(aq)) ----> 2 * H2O(l) (2 * ΔHrxn,1 = -2 * 56 kJ/mol H2O)
1 * (Zn(s) ----> Zn^2+(aq)) (1 * ΔHrxn,2 = -1 * 153.9 kJ)
2 * (1/2 H2(g) ----> H+(aq)) (2 * ΔHrxn,3 = 2 * 0 kJ)
Now, let's add up these modified reactions to obtain the overall reaction:
2 * (H+(aq) + OH-(aq)) + 1 * (Zn(s)) + 2 * (1/2 H2(g))
----> 2 * H2O(l) + Zn^2+(aq) + 2 * OH-(aq) + H2(g)
Simplifying:
2 * H+(aq) + 2 * OH-(aq) + Zn(s) + H2(g)
----> 2 * H2O(l) + Zn^2+(aq) + 2 * OH-(aq) + H2(g)
Now we can sum up the enthalpy changes for each reaction to calculate the enthalpy change for the overall reaction:
(2 * ΔHrxn,1) + ΔHrxn,2 + (2 * ΔHrxn,3) = (-2 * 56 kJ/mol H2O) + (-153.9 kJ) + (2 * 0 kJ)
Simplifying:
(-112 kJ/mol H2O) + (-153.9 kJ) + (0 kJ)
= -265.9 kJ/mol H2O
Therefore, the heat change for the slow reaction of zinc with water, using Hess's law, is -265.9 kJ/mol Zn.
To calculate the heat change for the slow reaction of zinc with water using Hess's law, we need to sum up the enthalpy changes of the given reactions to obtain the desired reaction.
First, let's assign the given reactions as follows:
Reaction 1: Zn(s) + 2 H2O(l) → Zn^2+(aq) + 2 OH^-(aq) + H2(g) (ΔHrxn = ?)
Reaction 2: H+(aq) + OH^-(aq) → H2O(l) (ΔHrxn,1 = -56 kJ/mol H2O)
Reaction 3: Zn(s) → Zn^2+(aq) (ΔHrxn,2 = -153.9 kJ)
Reaction 4: 1/2 H2(g) → H+(aq) (ΔHrxn,3 = 0.0 kJ)
Now, let's manipulate and combine these reactions to obtain the slow reaction:
1. Flip and multiply Reaction 1 by 2 to cancel out OH^-(aq):
2 Zn(s) + 4 H2O(l) → 2 Zn^2+(aq) + 4 OH^-(aq) + 2 H2(g)
2. Add Reaction 2 to cancel out OH^-(aq):
2 Zn(s) + 4 H2O(l) + H+(aq) + OH^-(aq) → 2 Zn^2+(aq) + 2 H2O(l) + H2(g)
3. Cancel out H2O(l) using Reaction 3:
2 Zn(s) + 4 H2O(l) + H+(aq) + OH^-(aq) + 2 H2(g) → 2 Zn^2+(aq) + 2 H2O(l) + H2(g)
4. Cancel out H2(g) using Reaction 4:
2 Zn(s) + 4 H2O(l) + H+(aq) + OH^-(aq) + 2 H2(g) → 2 Zn^2+(aq) + 2 H2O(l) + 2 H+(aq)
We can see that the slow reaction is now:
2 Zn(s) + 4 H2O(l) + H+(aq) + OH^-(aq) → 2 Zn^2+(aq) + 2 H2O(l) + 2 H+(aq)
Finally, let's sum up the enthalpy changes of the reactions to obtain the enthalpy change of the slow reaction:
ΔHrxn = ΔHrxn,2 + ΔHrxn,1 + ΔHrxn,3
Substituting the given values:
ΔHrxn = -153.9 kJ + (-56 kJ/mol H2O) + 0.0 kJ
Calculating the sum:
ΔHrxn = -209.9 kJ/mol Zn
Therefore, the heat change for the slow reaction of zinc with water is -209.9 kJ/mol Zn.