Questions: What quantity of heat (in kJ ) is absorbed in the process of making 0.781 mol of CF4 from the following reaction? C(s)+2 F2(g) → CF4(g) ΔH°=141.3 kJ / mol

Transcript text: What quantity of heat (in kJ ) is absorbed in the process of making 0.781 mol of $\mathrm{CF}_{4}$ from the following reaction? \[ C(\mathrm{~s})+2 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow C F_{4}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=141.3 \mathrm{~kJ} / \mathrm{mol} \]

Solution

Solution Steps

Step 1: Understand the Reaction and Given Data

The reaction provided is: \[ C(\mathrm{~s}) + 2 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow C F_{4}(\mathrm{~g}) \] The standard enthalpy change for this reaction, $\Delta \mathrm{H}^{\circ}$, is given as $141.3 \, \mathrm{kJ/mol}$. This means that 141.3 kJ of heat is absorbed for every mole of $\mathrm{CF}_{4}$ produced.

Step 2: Calculate the Heat Absorbed for 0.781 mol of $\mathrm{CF}_{4}$

To find the total heat absorbed when 0.781 mol of $\mathrm{CF}_{4}$ is produced, we use the formula: \[ \text{Heat absorbed} = \Delta \mathrm{H}^{\circ} \times \text{moles of } \mathrm{CF}_{4} \] Substituting the given values: \[ \text{Heat absorbed} = 141.3 \, \mathrm{kJ/mol} \times 0.781 \, \mathrm{mol} \]

Step 3: Perform the Calculation

Calculate the total heat absorbed: \[ \text{Heat absorbed} = 141.3 \times 0.781 = 110.4 \, \mathrm{kJ} \]