The standard Gibbs energy change for a reaction is`Delta G ^(@) =- 41.8 kJ mol ^(-1) ` at 700 K and 1 atm. Calculate the equilibrium constant of the reaction at 700 K.

Options

A. `1. 314 xx 10 ^(4)`
B. `3.431 xx 10 ^(4)`
C. `3.431 xx 10 ^(3)`
D. `1.314 xx 10 ^(3)`

Explanation

The standard Gibbs energy and equilibrium constant are related by the equation :
`Delta G ^(@) =- RT ln K = - 2. 303 RT log K`
Given that `Delta G ^(@) =- 41. 8 kJ =- 41800 J, T = 700 K and R = 8. 314JK ^(-1) mol ^(-1) .` Substiting the values in the equation , we get `- 41 800 =-2. 303 xx 8. 314 xx 700 log K`
`- 41 800 =- 134 0 3 log K or log K = ( 41 800)/( 13403) = 3. 1187 implies K = 1. 314 xx 10 ^(3)`