When H vap = 30 kJ/mol and S vap = 75 J mol^-1 K^-1, then the temperature of vapor, at one atmosphere, is:

Given: H vap = 30 kJ/mol and S vap = 75 J mol^-1 K^-1 at 1 atm. Find: The temperature of vaporization. At equilibrium for vaporization, use the Gibbs free energy relation: G = H - T S = 0 So, T = H vap S vap Convert enthalpy to joules: 30 kJ/mol = 30000 J/mol Substitute the values: T = 30000 J/mol75 J mol^-1 K^-1 = 400 K Therefore, the temperature of vaporization is 400 K. Given: H vap = 30 10^3 J/mol and S vap = 75 J mol^-1 K^-1. Find: The temperature. Since vaporization at the boiling point satisfies G = 0, the required temperature follows directly from: T = H S Now calculate: T = 30 10^375 = 400 K Therefore, the final answer is 400 K.

When H vap = 30 kJ/mol and S vap = 75 J mol^-1 K^-1, then the tem… — Solution | JEE Chemistry

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When $\Delta H_{\text{vap}} = 30 \, \text{kJ/mol}$ and $\Delta S_{\text{vap}} = 75 \, \text{J mol}^{-1}\text{K}^{-1}$ , then the temperature of vapor, at one atmosphere, is:

Answer

Correct answer:400

Step-by-step solution

Standard Method

Given: $\Delta H_{\text{vap}} = 30 \, \text{kJ/mol}$ and $\Delta S_{\text{vap}} = 75 \, \text{J mol}^{-1}\text{K}^{-1}$ at $1 \, \text{atm}$ .

Find: The temperature of vaporization.

At equilibrium for vaporization, use the Gibbs free energy relation:

\Delta G = \Delta H - T\Delta S = 0

So,

T = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}}

Convert enthalpy to joules:

30 \, \text{kJ/mol} = 30000 \, \text{J/mol}

Substitute the values:

T = \frac{30000 \, \text{J/mol}}{75 \, \text{J mol}^{-1}\text{K}^{-1}} = 400 \, \text{K}

Therefore, the temperature of vaporization is $400 \, \text{K}$ .

Direct Equilibrium Relation

Given: $\Delta H_{\text{vap}} = 30 \times 10^3 \, \text{J/mol}$ and $\Delta S_{\text{vap}} = 75 \, \text{J mol}^{-1}\text{K}^{-1}$ .

Find: The temperature.

Since vaporization at the boiling point satisfies $\Delta G = 0$ , the required temperature follows directly from:

T = \frac{\Delta H}{\Delta S}

Now calculate:

T = \frac{30 \times 10^3}{75} = 400 \, \text{K}

Therefore, the final answer is $400 \, \text{K}$ .

Common mistakes

Using $30$ directly instead of converting $30 \, \text{kJ/mol}$ into $30000 \, \text{J/mol}$ is wrong because $\Delta S$ is given in joules. Convert both quantities to consistent units before substitution.
Applying the relation $\Delta G = \Delta H - T\Delta S$ without setting $\Delta G = 0$ is incorrect here. At the vaporization equilibrium temperature, $\Delta G = 0$ , so use that condition first.
Multiplying by $\Delta S$ instead of dividing by it gives the wrong temperature. After rearranging, the correct expression is $T = \frac{\Delta H}{\Delta S}$ .

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