Certainly, let's analyze and solve this problem step-by-step.
### 1. Determine if the reaction is exothermic or endothermic:
Given reaction:
[tex]\[2 \text{Na} ( s ) + \text{Cl}_2 ( g ) \rightarrow 2 \text{NaCl} ( s )\][/tex]
The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction is -821.8 kJ. Because [tex]\(\Delta H\)[/tex] is negative, the reaction is exothermic. This means it releases heat.
### 2. Calculate the amount of heat transferred when 8.0 g of [tex]\( Na ( s ) \)[/tex] reacts:
- Molar mass of sodium (Na) = 23 g/mol.
- The enthalpy change is given for 2 moles of Na, which is -821.8 kJ.
First, we need to find the moles of Na:
[tex]\[\text{Moles of Na} = \frac{\text{mass of Na}}{\text{molar mass of Na}} = \frac{8.0 \text{ g}}{23 \text{ g/mol}} \approx 0.347826 \text{ mol}\][/tex]
Now, calculate the heat transferred:
Since [tex]\(-821.8 \text{ kJ}\)[/tex] is for 2 moles of Na, for 1 mole of Na, the heat transfer is [tex]\(\frac{-821.8}{2} \text{ kJ}\)[/tex].
[tex]\[\text{Heat transferred} = 0.347826 \text{ mol} \times \frac{-821.8 \text{ kJ}}{2 \text{ mol}} \approx -142.92 \text{ kJ}\][/tex]
### 3. How many grams of NaCl are produced during an enthalpy change of 10.0 kJ?
- Molar mass of NaCl = 58.44 g/mol.
- Given enthalpy change = 10.0 kJ.
First, calculate the moles of NaCl produced:
Since -821.8 kJ is for the production of 2 moles of NaCl, for 1 mole of NaCl, the heat is [tex]\(\frac{-821.8}{2} \text{ kJ}\)[/tex].
[tex]\[\text{Moles of NaCl produced} = \frac{10.0 \text{ kJ} \times 2 \text{ mol}}{-821.8 \text{ kJ}} \approx 0.02433 \text{ mol}\][/tex]
Now, calculate the mass of NaCl produced:
[tex]\[\text{Mass of NaCl} = 0.02433 \text{ mol} \times 58.44 \text{ g/mol} \approx 1.422 \text{ g}\][/tex]
### 4. How many kJ of heat are absorbed when 25.0 g of [tex]\( NaCl ( s ) \)[/tex] is decomposed into [tex]\( Na ( s ) \)[/tex] and [tex]\( Cl_2 ( g ) \)[/tex]:
- Molar mass of NaCl = 58.44 g/mol.
- Enthalpy change for the decomposition of NaCl is the reverse of formation, so it is also -821.8 kJ for decomposing 2 moles of NaCl.
First, find the moles of NaCl:
[tex]\[\text{Moles of NaCl} = \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{25.0 \text{ g}}{58.44 \text{ g/mol}} \approx 0.428 \text{ mol}\][/tex]
Then, calculate the heat absorbed:
[tex]\[\text{Heat absorbed} = 0.428 \text{ mol} \times \frac{-821.8 \text{ kJ}}{2 \text{ mol}} \approx -175.78 \text{ kJ}\][/tex]
Since heat absorbed should be a positive value for endothermic processes:
[tex]\[\text{Heat absorbed} \approx 175.78 \text{ kJ}\][/tex]
### Summary:
1. The reaction is exothermic.
2. When 8.0 g of Na reacts, approximately -142.92 kJ of heat is transferred.
3. Approximately 1.422 g of NaCl are produced during an enthalpy change of 10.0 kJ.
4. Approximately 175.78 kJ of heat are absorbed when 25.0 g of NaCl is decomposed.
