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Current Question (ID: 8323)
Question:
$\text{For the reaction}$
$\text{CH}_3\text{COOH (l)} + 2\text{O}_2 \text{(g)} \rightleftharpoons 2\text{CO}_2 \text{(g)} + 2\text{H}_2\text{O(l)}$
$\text{at } 25^{\circ}\text{C and 1 atm. pressure, } \Delta\text{H} = -874 \text{ kJ. The change in internal energy (}\Delta\text{E) is:}$
Options:
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1. $-874 \text{ kJ}$
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2. $-971.53 \text{ kJ}$
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3. $+971.53 \text{ kJ}$
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4. $+874 \text{ kJ}$
Solution:
$\text{Hint: } \Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}$
$\text{Step 1: Identify the relationship between enthalpy change and internal energy change}$
$\text{The relationship between enthalpy change (}\Delta\text{H) and internal energy change (}\Delta\text{E) is given by:}$
$\Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}$
$\text{where:}$
$\Delta\text{n}_g = \text{change in number of moles of gas in the reaction}$
$\text{R = universal gas constant = 8.314 J/(mol·K)}$
$\text{T = temperature in Kelvin}$
$\text{Step 2: Calculate the change in the number of moles of gas (}\Delta\text{n}_g\text{)}$
$\text{From the balanced equation:}$
$\text{CH}_3\text{COOH (l)} + 2\text{O}_2 \text{(g)} \rightleftharpoons 2\text{CO}_2 \text{(g)} + 2\text{H}_2\text{O(l)}$
$\text{Reactants: 2 moles of gas (}\text{O}_2\text{)}$
$\text{Products: 2 moles of gas (}\text{CO}_2\text{)}$
$\Delta\text{n}_g = \text{moles of gaseous products - moles of gaseous reactants}$
$\Delta\text{n}_g = 2 - 2 = 0$
$\text{Step 3: Calculate the internal energy change (}\Delta\text{E)}$
$\text{Since } \Delta\text{n}_g = 0\text{, from the equation } \Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}\text{, we get:}$
$\Delta\text{H} = \Delta\text{E} + 0 \times \text{RT}$
$\Delta\text{H} = \Delta\text{E}$
$\text{Given that } \Delta\text{H} = -874 \text{ kJ}$
$\text{Therefore, } \Delta\text{E} = -874 \text{ kJ}$
$\text{This makes physical sense because when there is no change in the number of moles of gas, the work of expansion or contraction is zero, so the enthalpy change equals the internal energy change.}$
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