madey21
Answered

Calculate the enthalpy, in [tex]kJ / mol[/tex], for the combustion of the hydrocarbon from the data from the previous steps.

[tex]\[
\begin{array}{c}
q_{\text{cal}} = 6191.78 \, \text{J} \quad n = 0.0019243682 \, \text{mol} \\
\Delta H = [?] \, \text{kJ} / \text{mol}
\end{array}
\][/tex]

Enter either a [tex]+[/tex] or [tex]-[/tex] sign AND the magnitude. Give your answer to three significant figures.



Answer :

Certainly! Here is the detailed, step-by-step solution for calculating the enthalpy change ([tex]\(\Delta H\)[/tex]) for the combustion of the hydrocarbon given the provided data.

### Step 1: Understand the given data
- Heat released ([tex]\(q_{\text{cal}}\)[/tex]) during combustion: 6191.78 Joules.
- Number of moles ([tex]\(n\)[/tex]) of the substance: 0.0019243682 moles.

### Step 2: Formula for Enthalpy change
The enthalpy change [tex]\(\Delta H\)[/tex] in kilojoules per mole ([tex]\(kJ/mol\)[/tex]) can be calculated using the formula:
[tex]\[ \Delta H = \frac{q_{\text{cal}}}{n} \][/tex]

### Step 3: Convert Joules to Kilojoules
Since 1 Joule = 0.001 Kilojoules, we first convert the heat released from Joules to Kilojoules:
[tex]\[ 6191.78 \, \text{J} = 6191.78 \times 0.001 \, \text{kJ} \][/tex]
[tex]\[ 6191.78 \, \text{J} = 6.19178 \, \text{kJ} \][/tex]

### Step 4: Calculate Enthalpy Change in [tex]\(kJ/mol\)[/tex]
Now, we divide the heat released (in [tex]\(kJ\)[/tex]) by the number of moles ([tex]\(n\)[/tex]):
[tex]\[ \Delta H = \frac{6.19178 \, \text{kJ}}{0.0019243682 \, \text{mol}} \][/tex]
[tex]\[ \Delta H = 3217.565 \, \text{kJ/mol} \][/tex]

### Step 5: Determine the sign of the Enthalpy Change
For combustion reactions, the enthalpy change is typically exothermic, so the sign of [tex]\(\Delta H\)[/tex] is negative:
[tex]\[ \Delta H = -3217.565 \, \text{kJ/mol} \][/tex]

### Step 6: Round to Three Significant Figures
Finally, we round the value to three significant figures:
[tex]\[ \Delta H = -3217.565 \approx -3218 \, \text{kJ/mol} \][/tex]

Therefore, the enthalpy change [tex]\(\Delta H\)[/tex] for the combustion of the hydrocarbon is [tex]\(-3217.565 \, kJ / mol\)[/tex], rounded to three significant figures, which is [tex]\(-3218 \, kJ / mol\)[/tex].