30.4 , kJ of heat is required to melt one mol | vedclass

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

Question

(A) The relationship between enthalpy change of fusion $(\Delta H_{fus})$,entropy change of fusion $(\Delta S_{fus})$,and melting point $(T_{mp})$ is

Vedclass · Accepted Answer

$1070$

Vedclass · Answer

(A) The relationship between enthalpy change of fusion $(\Delta H_{fus})$,entropy change of fusion $(\Delta S_{fus})$,and melting point $(T_{mp})$ is given by the formula: $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{mp}}$.Given: $\Delta H_{fus} = 30.4 \, kJ \, mol^{-1} = 30400 \, J \, mol^{-1}$ and $\Delta S_{fus} = 28.4 \, J \, K^{-1} \, mol^{-1}$.Substituting the values into the equation: $28.4 = \frac{30400}{T_{mp}}$.Solving for $T_{mp}$: $T_{mp} = \frac{30400}{28.4} \approx 1070.42 \, K$.Rounding to the nearest integer,the melting point is $1070 \, K$.

$30.4 \, kJ$ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is $28.4 \, J \, K^{-1} \, mol^{-1}$ at $1 \, atm$. The melting point of sodium chloride is $........... \, K$ (Nearest Integer).

Similar Questions

What will be $\Delta S$ for the reaction: $2A + 3B \to 4C + 5D$ (in $J \ K^{-1}$)?
Given:
$\Delta S_A^o = 100 \ J \ mol^{-1} \ K^{-1}$
$\Delta S_B^o = 120 \ J \ mol^{-1} \ K^{-1}$
$\Delta S_C^o = 200 \ J \ mol^{-1} \ K^{-1}$
$\Delta S_D^o = 150 \ J \ mol^{-1} \ K^{-1}$

What happens to the entropy of water when it freezes to form an ice cube?

For a spontaneous process,

When two gases are mixed,the entropy $......$

Calculate the entropy change when $1 \, \text{mole}$ of ice is converted into water at $0 \, ^\circ\text{C}$. (The enthalpy change for the conversion of ice to liquid at $0 \, ^\circ\text{C}$ is $6.0 \, \text{kJ} \, \text{mol}^{-1}$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module