Certainly! Let's go through each part of the problem step-by-step.
### (a) Thermochemical Equation for the Complete Combustion of Heptane
The chemical formula for heptane is \(\( \text{C}_7\text{H}_{16} \)\). The combustion of heptane in the presence of oxygen produces carbon dioxide (\(\(\text{CO}_2\)\)) and water (\(\(\text{H}_2\text{O}\)\)).
The balanced chemical equation for this reaction is:
\(\[ \text{C}_7\text{H}_{16} + 11\text{O}_2 \rightarrow 7\text{CO}_2 + 8\text{H}_2\text{O} \]\)
So, the thermochemical equation is:
\(\[ \text{C}_7\text{H}_{16} + 11\text{O}_2 \rightarrow 7\text{CO}_2 + 8\text{H}_2\text{O} \]\)
### (b) Enthalpy Change When 50 g of Heptane Are Burned
Given:
- The molar mass of heptane (\(\( \text{C}_7\text{H}_{16} \)\)) is 100.21 g/mol.
- The standard enthalpy change of combustion of heptane is \(\(-4817 \, \text{kJ/mol}\)\).
- Mass of heptane burned is 50 g.
First, we need to calculate the moles of heptane burned:
\(\[ \text{Moles of heptane burned} = \frac{\text{Mass of heptane}}{\text{Molar mass of heptane}} = \frac{50 \, \text{g}}{100.21 \, \text{g/mol}} \approx 0.499 \, \text{mol} \]\)
Next, we calculate the enthalpy change for burning 50 g of heptane:
\(\[ \text{Enthalpy change} = \text{Moles of heptane} \times \Delta H_{\text{comb}} = 0.499 \, \text{mol} \times (-4817 \, \text{kJ/mol}) \approx -2403.5 \, \text{kJ} \]\)
So, the enthalpy change when 50 g of heptane are burned is approximately \(\(-2403.5 \, \text{kJ}\)\).
### (c) Mass of Heptane Needed to Provide 100 MJ of Energy
Given:
- Desired energy: \(\(100 \, \text{MJ}\)\)
- Standard enthalpy change of combustion of heptane: \(\(-4817 \, \text{kJ/mol}\)\)
- 1 MJ = 1000 kJ
First, we convert the desired energy into kJ:
\(\[ 100 \, \text{MJ} = 100 \times 1000 \, \text{kJ} = 100000 \, \text{kJ} \]\)
Next, we calculate the moles of heptane needed to provide this amount of energy:
\(\[ \text{Moles needed} = \frac{\text{Desired energy}}{\Delta H_{\text{comb}}} = \frac{100000 \, \text{kJ}}{-4817 \, \text{kJ/mol}} \approx -20.802 \, \text{mol} \]\)
Since combustion is an exothermic reaction and the enthalpy change is negative, we take the absolute value of the moles needed:
\(\[ \text{Moles needed} \approx 20.802 \, \text{mol} \]\)
Finally, we determine the mass of heptane needed:
\(\[ \text{Mass of heptane needed} = \text{Moles needed} \times \text{Molar mass of heptane} = 20.802 \, \text{mol} \times 100.21 \, \text{g/mol} \approx 2080.3 \, \text{g} \]\)
Converting this mass to kilograms:
\(\[ 2080.3 \, \text{g} = 2.08 \, \text{kg} \]\)
So, the mass of heptane needed to provide 100 MJ of energy is approximately \(\(2.08 \, \text{kg}\)\).
