Question

Find the temperature at which the below reaction will be in equilibrium if the enthalpy and entropy change for the reaction is 30 kJ mol-1 and 105 J K-1 mol-1 respectively Br2(1) + Cl2(g) → 2BrCl(g)
(a) 273 K
(b) 300 K
(c) 450 K
(d) 295.7 K​

Answer 1

Explanation:

Correct option is

A

285.7K

Br

2

(l)+Cl

2

(g)⟶2BrCl(g)

△H=30KJ/mol=30×10

3

J/mol

△S=105J/K/mol

Since, △G=0, then only system is at equilibrium.

So, △G=△H−T△S

At equilibrium, △H=T△S

T

eq

=

△S

△H

=

105

30×10

3

=285.7K

Answer 2

Answer:

Explanation:

at equilibrium Gibbs free energy will be equal to zero, so,

we have a formula

ΔG = ΔH-TΔS

0 = 30*10⁻³ - T*105

Then,

T = 30*10⁻³/105

  = 285.71 K

maybe the option is given wrong, instead of giving 285.7K they might have given 295.7K