Due to the environmental concern of fluorocarbons as refrigerants, a refrigerant based on a mixture of hydrocarbons was used as a replacement. It is a patented blend of ethane, propane, butane, and isobutane. Isobutane has a normal boiling point of \(-12^{\circ} \mathrm{C}\). The molar specific heat of liquid phase and gas phase isobutane are $129.7 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\( and \)95.2 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ respectively. The heat of vaporization for this compound is $21.3 \mathrm{~kJ} / \mathrm{mol}\(. Calculate the heat required to convert \)25.0 \mathrm{~g}$ of isobutane from a liquid at \(-50^{\circ} \mathrm{C}\) to a gas at $40^{\circ} \mathrm{C}$.

First, we need to find the moles of isobutane given the mass: \(n = \frac{m}{M}\) where \(n\) represents the number of moles, \(m\) is the mass of isobutane, and \(M\) is the molar mass of isobutane. For isobutane, the molecular formula is \(C_4H_{10}\), and its molar mass is: \(M = 4 \times 12.01 + 10 \times 1.01 = 58.12 \mathrm{~g/mol}\) Now we can calculate the moles of isobutane: \(n = \frac{25.0 ~\mathrm{g}}{58.12 ~\mathrm{g/mol}} = 0.430 ~\mathrm{mol}\) Next, we need to calculate the heat required to heat the liquid from its initial temperature (-50°C) to the boiling point (-12°C). To do this, we will use the molar specific heat of the liquid phase and the formula: \(q_1 = nC_{p_{liquid}} \Delta T\) where \(q_1\) represents the heat required for this step, \(C_{p_{liquid}}\) is the molar specific heat of liquid isobutane, and \(\Delta T\) is the temperature change. \(\Delta T = T_{bp} - T_{initial} = -12 - (-50) = 38^{\circ} \mathrm{C}\) Now we can find the heat required: \(q_1 = (0.430 ~\mathrm{mol})(129.7 ~\mathrm{J/mol~K})(38~\mathrm{K}) = 2127 ~\mathrm{J}\)

Now we need to calculate the heat required for the phase change, vaporization. We will use the heat of vaporization and the formula: \(q_2 = n\Delta H_{vap}\) where \(q_2\) represents the heat required for this step, and \(\Delta H_{vap}\) is the heat of vaporization for isobutane. \( q_2 = (0.430 ~\mathrm{mol})(21.3 ~\mathrm{kJ/mol}) = 9.16 ~\mathrm{kJ}\) Convert to Joules: \(q_2 = 9160 ~\mathrm{J}\)

Finally, we need to calculate the heat required to heat the gas from the boiling point (-12°C) to the final temperature (40°C). To do this, we will use the molar specific heat of the gas phase and the formula: \(q_3 = nC_{p_{gas}}\Delta T\) where \(q_3\) represents the heat required for this step, \(C_{p_{gas}}\) is the molar specific heat of isobutane in the gas phase, and \(\Delta T\) is the temperature change. \(\Delta T = T_{final} - T_{bp} = 40 - (-12) = 52^{\circ} \mathrm{C}\) Now we can find the heat required: \(q_3 = (0.430 ~\mathrm{mol})(95.2 ~\mathrm{J/mol~K})(52 ~\mathrm{K}) = 2236 ~\mathrm{J}\)

To find the total heat required, we need to sum the heats calculated for the three steps: \(q_{total} = q_1 + q_2 + q_3\) \(q_{total} = 2127 ~\mathrm{J} + 9160 ~\mathrm{J} + 2236 ~\mathrm{J} = 13523 ~\mathrm{J}\) The heat required to convert 25.0 g of isobutane from a liquid at -50°C to a gas at 40°C is 13,523 J.