Question

5.000 g of compound x with molecular formula c4h6 are burned in a constant - pressure calorimeter containing 50.00 kg of water at 25°c. the temperature of the water is observed to rise by 1.065°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 3 significant digits.

Explanation:

Step1: Calculate the heat absorbed by water

The formula for heat absorbed \(Q = mc\Delta T\). Here, \(m = 50.00\space kg=50000\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(\Delta T=1.065^{\circ}C\).

$$Q = 50000\times4.18\times1.065$$

$$Q=50000\times4.18\times1.065 = 50000\times4.4517=222585\space J$$

Step2: Calculate the molar mass of \(C_4H_6\)

The molar mass of \(C_4H_6\): \(M=(4\times12.01 + 6\times1.008)\space g/mol=(48.04+6.048)\space g/mol = 54.088\space g/mol\)

Step3: Calculate the number of moles of \(C_4H_6\)

\(n=\frac{m}{M}\), where \(m = 5.000\space g\)

$$n=\frac{5.000}{54.088}\space mol\approx0.0924\space mol$$

Step4: Calculate the standard heat of formation

\(\Delta H=\frac{Q}{n}\) (Since heat released by reaction = heat absorbed by water)
\(\Delta H=\frac{- 222585\space J}{0.0924\space mol}\approx - 2.41\times10^{6}\space J/mol=-2410\space kJ/mol\)

Answer:

\(-2410\space kJ/mol\)