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Question Video: Calculating the Heat Released during the Condensation of Methanol Chemistry

How much heat in kilojoules, is released when 0.13 moles of methanol (g) at 64.7°C are converted to methanol (l)? Take the standard enthalpy of vaporization of methanol to be +35.2 kJ/mol. Give your answer to 2 decimal places.

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Video Transcript

How much heat in kilojoules is released when 0.13 moles of methanol gas at 64.7 degrees Celsius are converted to methanol liquid? Take the standard enthalpy of vaporization of methanol to be positive 35.2 kilojoules per mole. Give your answer to two decimal places.

In this question, we are told that a sample of methanol is converted from gas to liquid at a specific temperature. The name of the physical change that occurs when the gaseous form of a substance is transformed to the liquid form is condensation. We are told that during condensation, energy in the form of heat is released. Our job in answering this question is to determine how much heat is released when 0.13 moles of methanol gas condenses. The provided symbol Δ𝐻 vap represents the standard enthalpy of vaporization. The standard enthalpy of vaporization is the enthalpy change when one mole of a substance transforms from a liquid to a gas at standard temperature and pressure.

Vaporization, which is the conversion from liquid to gas, is the opposite of condensation. Energy must be absorbed by the liquid methanol in order for vaporization to happen. The given value of the standard enthalpy of vaporization for methanol is positive 35.2 kilojoules per mole. This means that if we had one mole of liquid methanol, it would need to absorb 35.2 kilojoules of energy to be fully converted to a gas. Processes that require the absorption of energy are endothermic, and their associated standard enthalpy values are positive.

To solve this problem, we’ll need to make use of the standard enthalpy of condensation, but we aren’t given this value. We already know that condensation is the opposite process of vaporization, so we can define the standard enthalpy of condensation as the enthalpy change when one mole of substance transforms from a gas to a liquid at standard conditions. The size of the standard enthalpy of condensation will be the same as the standard enthalpy of vaporization. However, the sign will be the opposite. Because heat is released during condensation, it’s classified as an exothermic process. We will need to use a negative sign when writing the value of the enthalpy of condensation. Let’s write negative 35.2 kilojoules per mole as the standard enthalpy of condensation for methanol.

If one mole of gaseous methanol releases 35.2 kilojoules of energy at standard conditions, enough energy will have been lost for all of the gas to be converted to a liquid. However, in this problem, the given amount of moles of methanol is quite a bit less, 0.13 moles. We’d expect less energy to be released during condensation. To determine the specific amount of energy released when 0.13 moles of methanol gas condenses at standard conditions, we’ll need to multiply our given amount of moles, 0.13, by the standard enthalpy for condensation of methanol, negative 35.2 kilojoules per mole. This gives us an answer of negative 4.576 kilojoules.

Finally, we’ll need to round our answer to two decimal places. Our final rounded answer is negative 4.58 kilojoules, which is the amount of heat released when 0.13 moles of methanol gas condenses at 64.7 degrees Celsius.

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