Question

a chemist measures the enthalpy change $\delta h$ during the following reaction:
\\2\\,\text{na}(s)\\; + \\;\text{cl}_2(g)\
ightarrow 2\\,\text{nacl}(s)\quad\delta h = -822.\\,\text{kj}\\

use this information to complete the table below. round each of your answers to the nearest kj.

reaction	$\delta h$

| $\frac{1}{2}\\,\text{nacl}(s)\\; \
ightarrow \\;\frac{1}{2}\\,\text{na}(s)\\; + \\;\frac{1}{4}\\,\text{cl}_2(g)$ | $\square\\,\text{kj}$ |
| $\text{na}(s)\\; + \\;\frac{1}{2}\\,\text{cl}_2(g)\\; \
ightarrow \\;\text{nacl}(s)$ | $\square\\,\text{kj}$ |
| $2\\,\text{nacl}(s)\\; \
ightarrow \\;2\\,\text{na}(s)\\; + \\;\text{cl}_2(g)$ | $\square\\,\text{kj}$ |

Explanation:

Step1: Analyze the original reaction

Original: $2\text{Na}(s) + \text{Cl}_2(g)
ightarrow 2\text{NaCl}(s)$, $\Delta H = -822\ \text{kJ}$

Step2: Calculate first reaction's ΔH

Reverse original: $2\text{NaCl}(s)
ightarrow 2\text{Na}(s) + \text{Cl}_2(g)$, $\Delta H = 822\ \text{kJ}$. Multiply by $\frac{1}{4}$: $\frac{1}{2}\text{NaCl}(s)
ightarrow \frac{1}{2}\text{Na}(s) + \frac{1}{4}\text{Cl}_2(g)$, $\Delta H = 822 \times \frac{1}{4} = 205.5 \approx 206\ \text{kJ}$

Step3: Calculate second reaction's ΔH

Divide original by 2: $\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g)
ightarrow \text{NaCl}(s)$, $\Delta H = -822 \times \frac{1}{2} = -411\ \text{kJ}$

Step4: Calculate third reaction's ΔH

Reverse original: $2\text{NaCl}(s)
ightarrow 2\text{Na}(s) + \text{Cl}_2(g)$, $\Delta H = 822\ \text{kJ}$

Answer:

reaction	ΔH

| $\frac{1}{2}\text{NaCl}(s)
ightarrow \frac{1}{2}\text{Na}(s) + \frac{1}{4}\text{Cl}_2(g)$ | 206 kJ |
| $\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g)
ightarrow \text{NaCl}(s)$ | -411 kJ |
| $2\text{NaCl}(s)
ightarrow 2\text{Na}(s) + \text{Cl}_2(g)$ | 822 kJ |