Question

For the following reaction
CCl4 (l) + H2 (g) → HCl (g) + CHCl3 (l)ΔrG° = – 103.7 kJ mol–1 at 298 K. Calculate the equilibrium constant for
this reaction

Answer 1

Its is written down below you can see

Answer 2

Answer:

The equilibrium constant for the reaction CCl₄ (l) + H₂ (g) → HCl (g) + CHCl₃ (l) is 1506577960458601000.

Given:

The value of ΔrG° = – 103.7 kJ mol–1 at

Temperature of 298 K.

To find:

The equilibrium constant for the reaction CCl₄ (l) + H₂ (g) → HCl (g) + CHCl₃ (l) at 298 K

Solution:

The equilibrium constant for the reaction at 298 K  can be calculated using the following equation:

K = e^(-ΔrG°/RT )

Where R is the universal gas constant (8.314 J/molK) and T is the temperature in Kelvin (298 K).

Plugging in the values, we get:

K = e^(-(-103.7 kJ/mol))/(0.008314 kJ/molK * 298 K)

K = e^(103.7/2.4472)

K = e^(41.85549)

K = 1506577960458601000

This can be interpreted as the ratio of the products to the reactants at equilibrium, which is 1506577960458601000 times higher than the starting concentration of reactants.

To know more about equilibrium, click below:

https://brainly.in/question/220726

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https://brainly.in/question/15035176

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