Question

given the data in exercise 115, calculate the energy that must be removed to convert 250. g of substance x from a gas at 100.°c to a solid at - 50.°c. assume x has a molar mass of 75.0 g/mol. substance x has the following properties: mp -15°c, bp 75°c, δhfus 5.0 kj/mol, δhvap 20. kj/mol, c(g) 1.0 j/g·°c, c(l) 2.5 j/g·°c, c(s) 3.0 j/g·°c, specific heat capacities

Explanation:

Step1: Calculate moles of substance X

The molar mass is not given in the problem - let's assume it's \(M\) g/mol. But we can calculate the number of moles \(n=\frac{m}{M}\), where \(m = 250.0\ g\). Since we are not using the molar - mass in the heat - calculation steps directly (as we will see), we can also work with the mass values.

Step2: Cool gas from 100°C to 75°C

Use the formula \(q_1 = mc_{(g)}\Delta T\), where \(m = 250.0\ g\), \(c_{(g)}=1.0\ J/g\cdot^{\circ}C\), and \(\Delta T=(75 - 100)^{\circ}C=- 25^{\circ}C\).
\[q_1=250.0\ g\times1.0\ J/g\cdot^{\circ}C\times(-25^{\circ}C)=-6250\ J=-6.25\ kJ\]

Step3: Condense gas to liquid

Use the formula \(q_2=-n\Delta H_{vap}\). First, find \(n=\frac{m}{M}\), but we can also use the fact that for the phase - change, \(q_2=-m\times\frac{\Delta H_{vap}}{M}\times M=-m\Delta H_{vap}\) (since \(n = \frac{m}{M}\)). Here, \(m = 250.0\ g\) and \(\Delta H_{vap}=20.0\ kJ/mol\). Let's assume \(M\) cancels out when we consider the mass - based calculation. \(q_2=-250.0\ g\times\frac{20.0\ kJ/mol}{M}\times M=-50000\ J=-50.0\ kJ\)

Step4: Cool liquid from 75°C to - 15°C

Use the formula \(q_3 = mc_{(l)}\Delta T\), where \(m = 250.0\ g\), \(c_{(l)} = 2.5\ J/g\cdot^{\circ}C\), and \(\Delta T=(-15 - 75)^{\circ}C=-90^{\circ}C\)
\[q_3=250.0\ g\times2.5\ J/g\cdot^{\circ}C\times(-90^{\circ}C)=-56250\ J=-56.25\ kJ\]

Step5: Freeze liquid to solid

Use the formula \(q_4=-n\Delta H_{fus}\). Similar to the vapor - liquid phase change, \(q_4=-m\Delta H_{fus}\), where \(m = 250.0\ g\) and \(\Delta H_{fus}=5.0\ kJ/mol\). \(q_4=-250.0\ g\times\frac{5.0\ kJ/mol}{M}\times M=-12500\ J=-12.5\ kJ\)

Step6: Cool solid from - 15°C to - 50°C

Use the formula \(q_5 = mc_{(s)}\Delta T\), where \(m = 250.0\ g\), \(c_{(s)}=3.0\ J/g\cdot^{\circ}C\), and \(\Delta T=(-50+15)^{\circ}C=-35^{\circ}C\)
\[q_5=250.0\ g\times3.0\ J/g\cdot^{\circ}C\times(-35^{\circ}C)=-26250\ J=-26.25\ kJ\]

Step7: Calculate total energy

\[q_{total}=q_1 + q_2+q_3+q_4+q_5\]
\[q_{total}=-6.25\ kJ-50.0\ kJ - 56.25\ kJ-12.5\ kJ-26.25\ kJ=-151.25\ kJ\]

Answer:

\(-151.25\ kJ\)