QuestionThe heat of the reaction, $\Delta H_{r \times n}$ , is found by dividing the $q_{r x n}$ by the moles of zinc used, $n_{z n}$ . $\Delta H_{r x n}=\frac{q_{r x n}}{n_{z n}}$
What is the $\Delta H_{r \times n}$ if $q_{r \times n}=-0.332 \mathrm{kcal}$ and $n=0.0084$ moles? a. $40 \mathrm{cal} / \mathrm{mol}$ b. $-40 . \mathrm{kcal} / \mathrm{mol}$ C. $40 \mathrm{kcal} / \mathrm{mol}$

Studdy Solution

STEP 1

1. The formula given is $\Delta H_{r \times n} = \frac{q_{r \times n}}{n_{z n}}$ .
2. We are provided with $q_{r \times n} = -0.332 \, \text{kcal}$ and $n_{z n} = 0.0084 \, \text{moles}$ .
3. We need to perform a division to find $\Delta H_{r \times n}$ .

STEP 2

1. Substitute the given values into the formula.
2. Perform the division to calculate $\Delta H_{r \times n}$ .
3. Interpret the result and match it with the given options.

STEP 3

Substitute the given values into the formula:
$\Delta H_{r \times n} = \frac{-0.332 \, \text{kcal}}{0.0084 \, \text{moles}}$

STEP 4

Perform the division to calculate $\Delta H_{r \times n}$ :
$\Delta H_{r \times n} = \frac{-0.332}{0.0084}$
Calculate the division:
$\Delta H_{r \times n} = -39.5238 \, \text{kcal/mol}$

STEP 5

Round the result to match the options provided. The closest option is:
$\Delta H_{r \times n} \approx -40 \, \text{kcal/mol}$
Therefore, the correct answer is:
b. $-40 \, \text{kcal/mol}$

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