Let's solve the problem step-by-step to find the enthalpy change (ΔH) for both reactions using the given formula:
[tex]\[
\Delta H = \frac{-q}{\text {moles}} \times \frac{\text{kJ}}{1000 \text{J}}
\][/tex]
where:
- [tex]\( q \)[/tex] is the heat energy in joules,
- "moles" is the number of moles of the substance.
### Reaction 1
1. Given data:
- [tex]\( q_{\text{reaction1}} = -30000 \)[/tex] J (energy in joules)
- moles of reactants for reaction 1 = 2
2. Convert energy from joules to kilojoules:
[tex]\[
q_{\text{reaction1 (kJ)}} = \frac{-30000}{1000} \text{ kJ} = -30 \text{ kJ}
\][/tex]
3. Calculate ΔH for reaction 1:
[tex]\[
\Delta H_{\text{reaction1}} = \frac{q_{\text{reaction1 (kJ)}}}{\text{moles reaction 1}} = \frac{-30}{2} = -15 \text{ kJ/mol}
\][/tex]
4. Round ΔH to 2 significant figures:
[tex]\[
\Delta H_{\text{reaction1}} \approx -15.0 \text{ kJ/mol}
\][/tex]
### Reaction 2
1. Given data:
- [tex]\( q_{\text{reaction2}} = -45000 \)[/tex] J (energy in joules)
- moles of reactants for reaction 2 = 3
2. Convert energy from joules to kilojoules:
[tex]\[
q_{\text{reaction2 (kJ)}} = \frac{-45000}{1000} \text{ kJ} = -45 \text{ kJ}
\][/tex]
3. Calculate ΔH for reaction 2:
[tex]\[
\Delta H_{\text{reaction2}} = \frac{q_{\text{reaction2 (kJ)}}}{\text{moles reaction 2}} = \frac{-45}{3} = -15 \text{ kJ/mol}
\][/tex]
4. Round ΔH to 2 significant figures:
[tex]\[
\Delta H_{\text{reaction2}} \approx -15.0 \text{ kJ/mol}
\][/tex]
### Summary:
- Reaction 1: [tex]\(-15.0\)[/tex] [tex]\( \text{kJ/mol} \)[/tex]
- Reaction 2: [tex]\(-15.0\)[/tex] [tex]\( \text{kJ/mol} \)[/tex]
So, the enthalpy changes for both reactions, recorded to 2 significant figures, are as follows:
- Reaction 1: [tex]\( -15.0 \text{ kJ/mol} \)[/tex]
- Reaction 2: [tex]\( -15.0 \text{ kJ/mol} \)[/tex]
