QuestionMay 28, 2025

What is the heat change when 4.72 g of Carbon reacts with excess O_(2) according to the following equation? Delta H=square kJ C+O_(2)arrow CO_(2) Delta H^circ =-393.5kJ "Report your answer with a positive sign (+) or negative sign (-) and to the appropriate number of significant figures. Is this reaction endothermic or exothermic? square

What is the heat change when 4.72 g of Carbon reacts with excess O_(2) according to the following equation? Delta H=square kJ C+O_(2)arrow CO_(2) Delta H^circ =-393.5kJ "Report your answer with a positive sign (+) or negative sign (-) and to the appropriate number of significant figures. Is this reaction endothermic or exothermic? square
What is the heat change when 4.72 g of Carbon reacts with excess O_(2) according to the following equation? Delta H=square kJ C+O_(2)arrow CO_(2) Delta H^circ =-393.5kJ "Report your answer with a positive sign (+) or negative sign (-) and to the appropriate number of significant figures. Is this reaction endothermic or exothermic? square

Solution
4.5(325 votes)

Answer

-154.7 kJ; Exothermic Explanation 1. Calculate moles of Carbon Molar mass of Carbon (C) is 12.01 g/mol. Moles of C = \frac{4.72 \, \text{g}}{12.01 \, \text{g/mol}} = 0.393 \, \text{mol}. 2. Calculate heat change Use the formula: Heat change = moles \times \Delta H^{\circ}. Here, \Delta H^{\circ} = -393.5 \, \text{kJ/mol}. So, Heat change = 0.393 \, \text{mol} \times (-393.5 \, \text{kJ/mol}) = -154.7 \, \text{kJ}. 3. Determine reaction type Since \Delta H is negative, the reaction is exothermic.

Explanation

1. Calculate moles of Carbon<br /> Molar mass of Carbon (C) is 12.01 g/mol. Moles of C = $\frac{4.72 \, \text{g}}{12.01 \, \text{g/mol}} = 0.393 \, \text{mol}$.<br /><br />2. Calculate heat change<br /> Use the formula: Heat change = moles $\times \Delta H^{\circ}$. Here, $\Delta H^{\circ} = -393.5 \, \text{kJ/mol}$. So, Heat change = $0.393 \, \text{mol} \times (-393.5 \, \text{kJ/mol}) = -154.7 \, \text{kJ}$.<br /><br />3. Determine reaction type<br /> Since $\Delta H$ is negative, the reaction is exothermic.
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