Question

use the standard reaction enthalpies given below to determine \\(\delta h^\circ_{\text{rxn}}\\) for the following reaction:

\\2\text{no(g)} + \text{o}_2\text{(g)} \
ightarrow 2\text{no}_2\text{(g)} \quad \delta h^\circ_{\text{rxn}} = ?\\

given:
\\\text{n}_2\text{(g)} + \text{o}_2\text{(g)} \
ightarrow 2\text{no(g)} \quad \delta h^\circ_{\text{rxn}} = +183\text{ kj}\\
\\1/2\text{n}_2\text{(g)} + \text{o}_2\text{(g)} \
ightarrow \text{no}_2\text{(g)} \quad \delta h^\circ_{\text{rxn}} = +33\text{ kj}\\

-150. kj
-117 kj
-333 kj
+115 kj
+238 kj

question 14
sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below. when 0.025 mol of na is added to 100.00 g of water, the temperature of the resulting solution rises from 25.00°c to 35.75°c. if the specific heat of the solution is \\(4.18\text{ j}/(\text{g} \cdot ^\circ\text{c})\\), calculate \\(\delta h\\) for the reaction, as written.

\\2\text{na(s)} + 2\text{h}_2\text{o(l)} \
ightarrow 2\text{naoh(aq)} + \text{h}_2\text{(g)} \quad \delta h = ?\\

-5.41 kj
-90.0 kj
-180 kj
-360 kj

Explanation:

Manipulate the given thermochemical equations

Using the Reaction Enthalpy Scaling knowledge point

We need to determine \(\Delta H^\circ_{\text{rxn}}\) for:

$$2\text{NO}(g) + \text{O}_2(g) ightarrow 2\text{NO}_2(g)$$

Given equations:

1. \(\text{N}_2(g) + \text{O}_2(g)

ightarrow 2\text{NO}(g) \quad \Delta H^\circ_1 = +183\text{ kJ}\)

1. \(\frac{1}{2}\text{N}_2(g) + \text{O}_2(g)

ightarrow \text{NO}_2(g) \quad \Delta H^\circ_2 = +33\text{ kJ}\)

We reverse equation (1) to place \(\text{NO}(g)\) on the reactant side:

$$2\text{NO}(g) ightarrow \text{N}_2(g) + \text{O}_2(g) \quad \Delta H^\circ_{\text{rev}} = -183\text{ kJ}$$

We multiply equation (2) by 2 to match the coefficient of \(\text{NO}_2(g)\):

$$\text{N}_2(g) + 2\text{O}_2(g) ightarrow 2\text{NO}_2(g) \quad \Delta H^\circ_{\text{mult}} = 2 \times (+33\text{ kJ}) = +66\text{ kJ}$$

Sum the modified equations

Using the Reaction Enthalpy Scaling knowledge point

Adding the two modified equations:

$$2\text{NO}(g) + \text{N}_2(g) + 2\text{O}_2(g) ightarrow \text{N}_2(g) + \text{O}_2(g) + 2\text{NO}_2(g)$$

Canceling common species on both sides (\(\text{N}_2\) and one \(\text{O}_2\)) yields:

$$2\text{NO}(g) + \text{O}_2(g) ightarrow 2\text{NO}_2(g)$$

The overall enthalpy change is:

$$\Delta H^\circ_{\text{rxn}} = -183\text{ kJ} + 66\text{ kJ} = -117\text{ kJ}$$

Calculate heat absorbed by the solution

Using the Specific Heat Capacity knowledge point

For Question 14, we first find the mass of the solution. Since \(0.025\text{ mol}\) of \(\text{Na}\) is added to \(100.00\text{ g}\) of water:

$$m_{\text{Na}} = 0.025\text{ mol} \times 22.99\text{ g/mol} \approx 0.57\text{ g}$$

$$m_{\text{soln}} = 100.00\text{ g} + 0.57\text{ g} = 100.57\text{ g}$$

The temperature change is:

$$\Delta T = 35.75^\circ\text{C} - 25.00^\circ\text{C} = 10.75^\circ\text{C}$$

The heat absorbed by the solution is:

$$q_{\text{soln}} = m_{\text{soln}} \times c \times \Delta T = 100.57\text{ g} \times 4.18\text{ J}/(\text{g}\cdot^\circ\text{C}) \times 10.75^\circ\text{C} \approx 4519\text{ J} = 4.519\text{ kJ}$$

Determine reaction enthalpy per mole of reaction

Using the Specific Heat Capacity knowledge point

Assuming an isolated system, the heat of the reaction is:

$$q_{\text{rxn}} = -q_{\text{soln}} = -4.519\text{ kJ}$$

This heat is released by the reaction of \(0.025\text{ mol}\) of \(\text{Na}\). The balanced equation is:

$$2\text{Na}(s) + 2\text{H}_2\text{O}(l) ightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$$

The stoichiometric coefficient of \(\text{Na}\) is 2. Therefore, the enthalpy change \(\Delta H\) for the reaction as written (per 2 moles of \(\text{Na}\)) is:

$$\Delta H = q_{\text{rxn}} \times \frac{2\text{ mol Na}}{0.025\text{ mol Na}} = -4.519\text{ kJ} \times 80 = -361.5\text{ kJ}$$

Rounding to two significant figures based on the \(0.025\text{ mol}\) of \(\text{Na}\) gives approximately \(-360\text{ kJ}\).

Answer:

Question 1

• o -150. kJ
• o -117 kJ (Correct answer)
• o -333 kJ
• o +115 kJ
• o +238 kJ

Question 14

• o -5.41 kJ
• o -90.0 kJ
• o -180 kJ
• o -360 kJ (Correct answer)