$\text{H}_2\text{O}_2(aq)\text{NaOCl}(aq)$, which is represented by the net-ionic equation shown abo — Thermodynamics Chemistry Question

The reaction is a redox reaction because the oxidation numbers of some atoms changed during the reaction (both oxygen and chlorine undergo changes in oxidation number).

1 point is earned for the correct answer along with a valid justification.

$$\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ$$
$$= -197\text{ kJ/mol}_{rxn} + (298\text{ K})\left(\frac{144\text{ J}}{\text{K}\cdot\text{mol}_{rxn}}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)$$
$$= -154\text{ kJ/mol}_{rxn}$$

1 point is earned for the correct calculation of $\Delta H^\circ$.

The temperature increases because the reaction is exothermic ($\Delta H^\circ < 0$).

1 point is earned for indicating an increase in temperature with a valid justification.

$$\Delta G^\circ = -RT \ln K$$
$$\ln K = \frac{-\Delta G^\circ}{RT} = \frac{-(-197,000\text{ J/mol})}{(8.314\text{ J/(mol}\cdot\text{K}))(298\text{ K})} = 79.5$$
$$K = e^{79.5} = 3 \times 10^{34}$$

1 point is earned for the correct value of $K$ with evidence of calculation.

$$PV = nRT$$
$$n = \frac{PV}{RT} = \frac{(0.988\text{ atm})(0.0400\text{ L})}{(0.08206\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(298\text{ K})} = 0.00162\text{ mol O}_2$$

$$0.00162\text{ mol O}_2 \times \frac{1\text{ mol H}_2\text{O}_2}{1\text{ mol O}_2} = 0.00162\text{ mol H}_2\text{O}_2\text{ needed}$$

$$0.00162\text{ mol H}_2\text{O}_2 \times \frac{\text{L}}{0.800\text{ mol H}_2\text{O}_2} = 0.00202\text{ L}$$

1 point is earned for calculating the number of moles of $\text{O}_2$ needed.
1 point is earned for calculating the volume of the $\text{H}_2\text{O}_2$ solution that should be added.

36.5 mL (values within ± 0.4 of 36.5 are acceptable)

1 point is earned for the correct reading of the meniscus level to three significant figures. || $$\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{36.5\text{ mL}}{40.0\text{ mL}} \times 100 = 91.3\%$$

1 point is earned for the correct percent yield. || No, the gas also contains water vapor and air that was originally in the flask.

1 point is earned for the correct answer with a valid explanation. (Only one of the two extra gases is required for the point.)

Disagree. The very large value of $K$ implies that the reaction goes essentially to completion, so essentially all of the $\text{H}_2\text{O}_2$ reacts to form $\text{O}_2$.

1 point is earned for an appropriate conclusion and valid justification based on the value of $K$ in part (d).