Question

2.000 g of compound x with molecular formula c₄h₆ are burned in a constant - pressure calorimeter containing 20.00 kg of water at 25 °c. the temperature of the water is observed to rise by 1.081 °c. (you may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) calculate the standard heat of formation of compound x at 25 °c. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Explanation:

Step1: Calculate heat absorbed by water

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m = 20.00\space kg = 20000\space g \), \( c = 4.184\space J/g^\circ C \), \( \Delta T = 1.081^\circ C \).
\( q = 20000\space g \times 4.184\space J/g^\circ C \times 1.081^\circ C \)
\( q = 20000 \times 4.184 \times 1.081 \space J \)
\( q = 20000 \times 4.522904 \space J \)
\( q = 90458.08\space J = 90.45808\space kJ \)

Step2: Calculate moles of Compound X (\( C_4H_6 \))

Molar mass of \( C_4H_6 \): \( 4 \times 12.01 + 6 \times 1.008 = 48.04 + 6.048 = 54.088\space g/mol \)
Moles \( n = \frac{2.000\space g}{54.088\space g/mol} \approx 0.036976\space mol \)

Step3: Determine heat of combustion (\( \Delta H_{combustion} \))

Heat released by combustion is equal to heat absorbed by water (with sign: combustion is exothermic, so \( \Delta H_{combustion} = -\frac{q}{n} \))
\( \Delta H_{combustion} = -\frac{90.45808\space kJ}{0.036976\space mol} \approx -2446.3\space kJ/mol \)

Step4: Write combustion reaction and use Hess's law

Combustion reaction: \( C_4H_6(g) + \frac{11}{2}O_2(g)
ightarrow 4CO_2(g) + 3H_2O(l) \)
Standard heats of formation: \( \Delta H_f(CO_2) = -393.5\space kJ/mol \), \( \Delta H_f(H_2O) = -285.8\space kJ/mol \), \( \Delta H_f(O_2) = 0\space kJ/mol \)
\( \Delta H_{combustion} = \sum \Delta H_f(products) - \sum \Delta H_f(reactants) \)
\( -2446.3 = [4 \times (-393.5) + 3 \times (-285.8)] - [\Delta H_f(C_4H_6) + \frac{11}{2} \times 0] \)
Calculate products: \( 4 \times (-393.5) = -1574 \), \( 3 \times (-285.8) = -857.4 \); sum: \( -1574 - 857.4 = -2431.4\space kJ/mol \)
So, \( -2446.3 = -2431.4 - \Delta H_f(C_4H_6) \)
\( \Delta H_f(C_4H_6) = -2431.4 + 2446.3 = 14.9\space kJ/mol \)

Answer:

\( \boldsymbol{15.0\space kJ/mol} \) (rounded to correct significant figures, or precise value from calculation)