One mole of NaCl (s) on melting absorbed 30.5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) At the melting point,the process of melting is in equilibrium,so $\Delta G = 0$. From the relation $\Delta G = \Delta H - T\Delta S$,we get $T = \

Vedclass · Accepted Answer

$1059$

Vedclass · Answer

(A) At the melting point,the process of melting is in equilibrium,so $\Delta G = 0$. From the relation $\Delta G = \Delta H - T\Delta S$,we get $T = \frac{\Delta H}{\Delta S}$. Given: $\Delta H = 30.5 \ kJ \ mol^{-1} = 30500 \ J \ mol^{-1}$ and $\Delta S = 28.8 \ J \ K^{-1} \ mol^{-1}$. Substituting the values: $T = \frac{30500 \ J \ mol^{-1}}{28.8 \ J \ K^{-1} \ mol^{-1}} \approx 1059 \ K$.

One mole of $NaCl$ $(s)$ on melting absorbed $30.5 \ kJ$ of heat and its entropy increased by $28.8 \ J \ K^{-1}$. The melting point of $NaCl$ is ............. $K$.

Similar Questions

Why does the entropy of a substance increase with an increase in temperature?

In which of the following reactions is $\Delta S$ maximum?

In which of the following processes is $\Delta S$ negative?

The entropy change for the expansion of $2 \, \text{mol}$ of an ideal gas from $2 \, \text{L}$ to $20 \, \text{L}$ at a temperature of $27 \, ^oC$ is ...... $(R = 2 \, \text{cal/mol K})$

Which state functions are required to understand the second law of thermodynamics?

Vedclass Test Series

Exam Paper Generator

Online Exam Module