Question

9.000 g of compound x with molecular formula c4h8 are burned in a constant - pressure calorimeter containing 25.00 kg of water at 25°c. the temperature of the water is observed to rise by 3.940°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.

Explanation:

Step1: Calculate the heat absorbed by water

Use the formula \(Q = mc\Delta T\). Here, \(m = 25.00\space kg=25000\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(\Delta T=3.940^{\circ}C\).

$$Q = 25000\times4.18\times3.940$$

$$Q = 25000\times16.4692$$

$$Q = 411730\space J$$

Step2: Calculate the number of moles of compound \(X\)

The molar mass of \(C_4H_8\) is \(M=(4\times12 + 8\times1)\space g/mol=56\space g/mol\). The mass of compound \(X\) is \(m = 9.000\space g\). The number of moles \(n=\frac{m}{M}\), so \(n=\frac{9.000}{56}\space mol\approx0.1607\space mol\)

Step3: Calculate the standard heat of formation

The combustion reaction is \(C_4H_8+6O_2
ightarrow4CO_2 + 4H_2O\). The heat of combustion \(\Delta H_{comb}\) (opposite of heat absorbed by water since heat released by combustion is absorbed by water) for \(n = 0.1607\space mol\) is \(\Delta H=- 411730\space J=-411.73\space kJ\).
For \(1\space mol\) of \(C_4H_8\), \(\Delta H_{comb}=\frac{-411.73}{0.1607}\space kJ/mol\approx - 2562\space kJ/mol\)
Using the formula \(\Delta H_{comb}=\sum n\Delta H_f(products)-\sum m\Delta H_f(reactants)\)
\(\Delta H_f(CO_2)=- 393.5\space kJ/mol\), \(\Delta H_f(H_2O(l))=-285.8\space kJ/mol\), \(\Delta H_f(O_2) = 0\space kJ/mol\)
\(\Delta H_{comb}=4\times(-393.5)+4\times(-285.8)-\Delta H_f(C_4H_8)\)
\(-2562=4\times(-393.5)-4\times285.8-\Delta H_f(C_4H_8)\)
\(-2562=-1574-1143.2-\Delta H_f(C_4H_8)\)
\(-2562=-2717.2-\Delta H_f(C_4H_8)\)
\(\Delta H_f(C_4H_8)=- 155\space kJ/mol\)

Answer:

\(-155\space kJ/mol\)