Question

10.00 g of compound x with molecular formula c₄h₈ are burned in a constant - pressure calorimeter containing 10.00 kg of water at 25°c. the temperature of the water is observed to rise by 10.94°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 2 significant digits.
281 kj/mol

Explanation:

Step1: Calculate the heat absorbed by water

The formula for heat \(q = mc\Delta T\). Here, \(m = 10.00\space kg=10000\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(\Delta T=10.94^{\circ}C\).
\(q = 10000\times4.18\times10.94\) \(= 457292\space J = 457.292\space kJ\)

Step2: Calculate the number of moles of \(C_4H_8\)

The molar mass of \(C_4H_8\) is \(M=(4\times12 + 8\times1)\space g/mol=56\space g/mol\). The number of moles \(n=\frac{m}{M}\), where \(m = 10.00\space g\). So \(n=\frac{10.00}{56}\space mol\approx0.1786\space mol\)

Step3: Calculate the heat of combustion per mole

The heat of combustion \(\Delta H_{comb}\) (negative because combustion is exothermic). Since \(q = 457.292\space kJ\) is absorbed by water (heat released by combustion of \(C_4H_8\) is \(q_{rxn}=- 457.292\space kJ\)). \(\Delta H_{comb}=\frac{q_{rxn}}{n}\)
\(\Delta H_{comb}=\frac{- 457.292}{0.1786}\space kJ/mol\approx - 2560\space kJ/mol\)

Step4: Use the standard heat of formation formula

The combustion reaction of \(C_4H_8\): \(C_4H_8(g)+6O_2(g)
ightarrow4CO_2(g)+4H_2O(l)\)
The standard heat of formation formula \(\Delta H_{comb}=\sum n\Delta H_f^{\circ}(products)-\sum n\Delta H_f^{\circ}(reactants)\)
We know \(\Delta H_f^{\circ}(O_2,g) = 0\space kJ/mol\), \(\Delta H_f^{\circ}(CO_2,g)=- 393.5\space kJ/mol\), \(\Delta H_f^{\circ}(H_2O,l)=- 285.8\space kJ/mol\)
\(\Delta H_{comb}=4\times(-393.5)+4\times(- 285.8)-\Delta H_f^{\circ}(C_4H_8,g)\)
\(-2560=4\times(-393.5)-4\times285.8-\Delta H_f^{\circ}(C_4H_8,g)\)
\(-2560=-1574 - 1143.2-\Delta H_f^{\circ}(C_4H_8,g)\)
\(\Delta H_f^{\circ}(C_4H_8,g)=(-1574 - 1143.2 + 2560)\space kJ/mol\)
\(\Delta H_f^{\circ}(C_4H_8,g)= - 157.2\space kJ/mol\approx - 160\space kJ/mol\)

Answer:

\(-160\space kJ/mol\)