CH 221-Ziegler Chapter 5 Worksheet Enthalpy, showing solved problems on heat calculation, reaction energy, and energy diagrams.

Here are the step-by-step solutions for the problems on the worksheet.

1. Heat Absorbed by Aluminum

Goal: Calculate the heat energy ($q$) needed to raise the temperature of aluminum.

* Step 1: Find the change in temperature ($\Delta T$).
Subtract the starting temperature from the final temperature.
$$26.8^\circ\text{C} - 25.0^\circ\text{C} = 1.8^\circ\text{C}$$

* Step 2: Use the specific heat formula.
The formula is $q = m \cdot c \cdot \Delta T$.
* Mass ($m$) = $4.75 \text{ g}$
* Specific heat ($c$) = $0.902 \text{ J/g}^\circ\text{C}$
* Change in temp ($\Delta T$) = $1.8^\circ\text{C}$

* Step 3: Calculate.
$$q = 4.75 \times 0.902 \times 1.8$$
$$q = 7.7121 \text{ J}$$

Rounding to two significant figures (based on the temperature difference), we get 7.7 J.

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2. Reaction of Copper(II) Oxide

Reaction: $2\text{CuO}(s) \rightarrow 2\text{Cu}(s) + \text{O}_2(g)$ with $\Delta H = 312.2 \text{ kJ}$.

Part a: Heat involved if 20.0 g Cu is formed.

* Step 1: Convert grams of Cu to moles.
The molar mass of Copper (Cu) is approximately $63.55 \text{ g/mol}$.
$$\text{Moles of Cu} = \frac{20.0 \text{ g}}{63.55 \text{ g/mol}} \approx 0.3147 \text{ mol}$$

* Step 2: Use the mole ratio to find energy.
According to the equation, forming 2 moles of Cu requires 312.2 kJ.
$$\text{Energy} = 0.3147 \text{ mol Cu} \times \frac{312.2 \text{ kJ}}{2 \text{ mol Cu}}$$
$$\text{Energy} = 0.3147 \times 156.1$$
$$\text{Energy} \approx 49.12 \text{ kJ}$$

Rounding to 3 significant figures, the answer is 49.1 kJ (or 49.2 kJ depending on precise atomic mass used).

Part b: Grams of CuO consumed if 156 kJ of heat is produced.
*(Note: The problem likely implies "involved" or "absorbed" since the reaction is endothermic, but we will follow the math provided).*

* Step 1: Find moles of reaction based on energy.
We know $312.2 \text{ kJ}$ corresponds to reacting 2 moles of CuO.
$$\text{Moles of CuO} = 156 \text{ kJ} \times \frac{2 \text{ mol CuO}}{312.2 \text{ kJ}}$$
$$\text{Moles of CuO} \approx 0.9994 \text{ mol}$$

* Step 2: Convert moles of CuO to grams.
Molar mass of CuO = $63.55 (\text{Cu}) + 16.00 (\text{O}) = 79.55 \text{ g/mol}$.
$$\text{Mass} = 0.9994 \text{ mol} \times 79.55 \text{ g/mol}$$
$$\text{Mass} \approx 79.5 \text{ g}$$

The calculated answer is approximately 79.5 g CuO.

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3. Endothermic Reactions

a. Sign of $\Delta H$:
Endothermic reactions absorb heat from the surroundings. Because the system gains energy, the change in enthalpy ($\Delta H$) is positive.
Answer: Positive (+)

b. Feel of the container:
Since the reaction takes heat *in* from the surroundings to make the reaction happen, it pulls heat away from your hand if you touch the container.
Answer: Cold

c. Energy Diagram:
For an endothermic reaction, the products have higher energy than the reactants because energy was absorbed.
Description: The line starts low on the left (Reactants) and goes up to a higher level on the right (Products).

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4. Platinum Catalyst Reaction

Reaction: $4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$; $\Delta H = -609 \text{ kJ}$.

Part a: Enthalpy change for the reverse reaction.
The new equation is exactly the reverse of the original one. When you flip a chemical equation, you flip the sign of $\Delta H$.
Original: $-609 \text{ kJ}$
Reverse: $+609 \text{ kJ}$
Answer: $\Delta H = +609 \text{ kJ}$

Part b: Enthalpy change when 10.0 g $\text{NH}_3$ reacts.

* Step 1: Convert grams of $\text{NH}_3$ to moles.
Molar mass of $\text{NH}_3$ = $14.01 (\text{N}) + 3 \times 1.008 (\text{H}) \approx 17.03 \text{ g/mol}$.
$$\text{Moles} = \frac{10.0 \text{ g}}{17.03 \text{ g/mol}} \approx 0.5872 \text{ mol}$$

* Step 2: Calculate energy using the mole ratio.
The equation shows that reacting 4 moles of $\text{NH}_3$ releases 609 kJ (indicated by the negative sign).
$$\text{Energy} = 0.5872 \text{ mol NH}_3 \times \frac{-609 \text{ kJ}}{4 \text{ mol NH}_3}$$
$$\text{Energy} = 0.5872 \times -152.25$$
$$\text{Energy} \approx -89.4 \text{ kJ}$$

The calculated value is approximately -89.4 kJ.

Final Answer:
1. 7.7 J
2a. 49.1 kJ
2b. 79.5 g CuO
3a. Positive (+)
3b. Cold
4a. +609 kJ
4b. -89.4 kJ