Question

49,936.64 j of energy are released when 0.015475 mol c₆h₅cooh react. what is the enthalpy of combustion for benzoic acid? δhᵣₓₙ = ? (kj/mol) enter either a + or - sign and the magnitude. give your answer to three significant figures.

Explanation:

Step1: Convert energy to kJ

Given energy is 49936.64 J. Since 1 kJ = 1000 J, we convert J to kJ:
$49936.64\ \text{J} \times \frac{1\ \text{kJ}}{1000\ \text{J}} = 49.93664\ \text{kJ}$

Step2: Calculate enthalpy of combustion

Enthalpy of combustion ($\Delta H_{\text{rxn}}$) is energy released per mole. The moles of $\ce{C6H5COOH}$ is 0.015475 mol. Since energy is released, $\Delta H$ is negative.
$\Delta H_{\text{rxn}} = \frac{-49.93664\ \text{kJ}}{0.015475\ \text{mol}}$

Calculate the value:
$\frac{49.93664}{0.015475} \approx 3227$ (so with sign: $-3230$ when rounded to three significant figures? Wait, let's do the division properly:

$49.93664 \div 0.015475 = \frac{49.93664}{0.015475} \approx 3227.0$

Wait, 49.93664 / 0.015475: let's compute 49.93664 ÷ 0.015475. Let's see, 0.015475 × 3227 = 0.015475×3000=46.425, 0.015475×227=0.015475×200=3.095, 0.015475×27=0.417825; total 46.425+3.095=49.52+0.417825=49.937825, which is very close to 49.93664. So the value is approximately 3227, but we need three significant figures. Wait, 49.93664 has five significant figures, 0.015475 has five. When dividing, the result should have the same number of significant figures as the least precise? Wait, no: 49.93664 (J converted to kJ: 49.93664 kJ, which is five sig figs) divided by 0.015475 mol (five sig figs). So the result should have five, but the question says three significant figures. So 3227 rounded to three significant figures is 3230? Wait, 3227: the first three digits are 3,2,2; the next digit is 7, which is more than 5, so we round up the third digit: 3230? Wait, no: 3227. The number is 3.227 × 10³. To three significant figures, it's 3.23 × 10³, which is 3230. But let's check the calculation again.

Wait, 49936.64 J is released, so the energy change is -49936.64 J. Convert to kJ: -49.93664 kJ. Then divide by moles: -49.93664 kJ / 0.015475 mol. Let's compute that:

-49.93664 / 0.015475 = - (49.93664 / 0.015475) ≈ -3227. So to three significant figures, 3230? Wait, 3227 rounded to three significant figures: the third digit is 2, the next digit is 7, so we round up the 2 to 3, making it 3230. Wait, 3227: 3 (1st), 2 (2nd), 2 (3rd), 7 (4th). So rounding the third digit: 2 +1 =3, so 3230. So -3230 kJ/mol? Wait, no, 3227 is approximately 3.23 × 10³, so -3.23 × 10³ kJ/mol, which is -3230 kJ/mol? Wait, no, 3227 to three significant figures is 3230? Wait, 3227: the first three significant figures are 3,2,2. The fourth is 7, which is greater than 5, so we round the third digit up: 3230. Yes. So the enthalpy of combustion is -3230 kJ/mol? Wait, no, wait: 49.93664 / 0.015475 = let's do the division more accurately. 0.015475 × 3227 = 0.015475 * 3227. Let's compute 3227 * 0.015475:

3227 * 0.01 = 32.27
3227 * 0.005 = 16.135
3227 * 0.0004 = 1.2908
3227 * 0.00007 = 0.22589
3227 * 0.000005 = 0.016135

Wait, maybe better to do 49.93664 ÷ 0.015475. Let's multiply numerator and denominator by 1e6 to eliminate decimals: 49936640 ÷ 15475. Let's divide 49936640 by 15475.

15475 × 3227 = 15475 × 3000 = 46,425,000; 15475 × 227 = 15475 × 200 = 3,095,000; 15475 × 27 = 417,825; so 46,425,000 + 3,095,000 = 49,520,000 + 417,825 = 49,937,825. Which is very close to 49,936,640. The difference is 49,937,825 - 49,936,640 = 1,185. So 3227 - (1,185 / 15,475) ≈ 3227 - 0.0766 ≈ 3226.9234. So approximately 3227. So to three significant figures, 3230? Wait, 3227: the first three digits are 3,2,2. The fourth digit is 7, which is more than 5, so we round the third digit up: 3230. So the enthalpy of combustion is -3230 kJ/mol? Wait, but let's check the calculation again.

Wait, the energy released is 49,936.64 J, so the enthalpy change is negative (since combustion releases energy). So:

$\Delta H_{\text{rxn}} = \frac{-49936.64\ \text{J}}{0.015475\ \text{mol}} \times \frac{1\ \text{kJ}}{1000\ \text{J}}$

Wait, no: first convert J to kJ: 49936.64 J = 49.93664 kJ. Then divide by moles:

$\Delta H_{\text{rxn}} = \frac{-49.93664\ \text{kJ}}{0.015475\ \text{mol}} \approx -3227\ \text{kJ/mol}$

Now, round to three significant figures: 3227 rounded to three significant figures is 3230? Wait, 3227: the number is 3.227 × 10³. To three significant figures, it's 3.23 × 10³, which is 3230. So the answer is -3230 kJ/mol? Wait, but let's check with exact calculation:

49.93664 ÷ 0.015475 = 49.93664 / 0.015475 ≈ 3227. So three significant figures: 3230 (since the fourth digit is 7, which rounds up the third digit from 2 to 3). So the enthalpy of combustion is -3230 kJ/mol? Wait, no, 3227 is approximately 3.23 × 10³, so -3.23 × 10³ kJ/mol, which is -3230 kJ/mol. Wait, but maybe I made a mistake in significant figures. Let's see: 49936.64 J has six significant figures, 0.015475 mol has five. When dividing, the result should have five significant figures, but the question asks for three. So 3227 rounded to three is 3230. Yes.

Answer:

-3230