Question

One mole of NaCl on melting absorbed 30.5KJ of heat and it's entropy is increased by 28.8Jk.The melting point of NaCl is

Answer 1

Answer : The melting point of NaCl is 786 °C

Explanation :

Gibb's free energy ΔG , heat absorbed ΔH and entropy ΔS are related to each other by following equation.

$\bigtriangleup G = \bigtriangleup H - T \times \bigtriangleup S$

During melting , the temperature of the substance remains constant because the reaction is in equilibrium.

Therefore, during melting ΔG = 0.

Let us plug in this value of ΔG in above equation.

$0 = \bigtriangleup H - T \times \bigtriangleup S$

$\bigtriangleup H = T \bigtriangleup S$ ...........equation (1)

Here T is the melting temperature also known as melting point temperature.

We have been given ΔH = 30.5 kJ. Let us convert this to J

$30.5 kJ \times\frac{1000 J}{1 kJ}  = 30500 J$

ΔH = 30500 J

ΔS = 28.8 J/K

Let us plug in these values in equation (1)

$30500 J = T \times 28.8 J/K$

$T =\frac{30500 J}{28.8 J/K}$

$T = 1059 K$

Melting point of NaCl is 1059 K - 273 = 786 °C

Answer 2

Answer:

Explanation:as the delta g is zero becoz if equilibrum according to the formula of gibbs energy it is 0 hence delta h is equal to T delta s so by putting value we get 30500/28.8=1059K