madey21
Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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.

Sagot :

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].