Question

Calculate the heat flow when 15.6 G solid benzene at 0.0 degrees C is converted to liquid benzene at 5.5 degrees Celsius

Answer 1

Answer on Question #84726 – Chemistry – Other

Task:

Calculate the heat flow when 15.6 g solid benzene at 0.0°C is converted to liquid benzene at 5.5°C.

Solution:

Benzene freezes at 5.5°C = 278.65 K (melting point of benzene).

Heat capacity for solid benzene = 118.4 J/mol*K.

Enthalpy change of fusion for benzene = 9.9 kJ/mol.

Molar mass of benzene = 78.11 g/mol.

Convert these values from J/mol*K to J/K:

Heat capacity for solid benzene = C(C₆H₆, solid) = (118.4 J mol⁻¹ K⁻¹) / (78.11 g mol⁻¹) = 1.516 J/K = 1.516 J/°C;

Heat capacity for solid benzene = 1.516 J/°C.

Enthalpy change of fusion for benzene = ΔfusH° = (9900 J mol⁻¹) / (78.11 g mol⁻¹) = 126.744 J/g;

Enthalpy change of fusion for benzene = 126.744 J/g.

Solid benzene will release some of its internal energy to the surroundings when it heats and then melted.

1) heating the solid benzene from 0°C to 5.5°C;

2) melting the benzene.

1) Heating solid benzene:

$$\Delta Q = C m \Delta T$$ $$\Delta Q _ {1} = C m \Delta T = (1.516 \mathrm{J/g} \cdot ^ {\circ} \mathrm{C}) * (15.6 \mathrm{g}) * (5.5^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) = 130.073 \mathrm{J}$$ $$\Delta Q _ {1} = 130.073 \mathrm{J}$$

2) Melting the solid benzene:

$$\Delta Q = m ^ {*} \Delta_ {f u s} H ^ {o}$$ $$\Delta Q _ {2} = m ^ {*} \Delta_ {f u s} H ^ {o} = (15.6 \mathrm {g}) ^ {*} (126.744 \mathrm {J / g}) = 1977.206 \mathrm {J}$$ $$\Delta Q _ {2} = 1977.206 \mathrm {J}$$

Finally, the total energy released is:

$$\Delta Q = \Delta Q _ {1} + \Delta Q _ {2} = 130.073 \mathrm {J} + 1977.206 \mathrm {J} = 2107.279 \mathrm {J}$$ $$\Delta Q = 2107.279 \mathrm {J} \approx 2.1 \mathrm {kJ}$$

Answer: Heat flow = 2.1 kJ.

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