Question

Calculate the free energy δg at 25 ∘c for the nonstandard conditions at point b where the reaction quotient q is 2.75×10−5.

Answer 1

Answer: ctrl+c paste= ctrl=v

Step-by-step explanation:

Answer 2

This is an incomplete question, here is a complete question.

Calculate the free energy ΔG at 25 °C for the nonstandard conditions at point B where the reaction quotient Q is 2.75 × 10⁻⁵.

In solving Part A, we found that ΔG = -40.82 kJ/mol

Answer : The value of
`\Delta G` is, -66.8 kJ/mol

Explanation :

The expression for free energy is:

`\Delta G=\Delta G^o+RT\ln Q`

where,

`\Delta G` = free energy = ?

`\Delta G_^o` = standard Gibbs free energy = -40.82 kJ/mol

R = gas constant =
`8.314* 10^(-3)kJ/mole.K`

T = temperature = 298 K

Q = equilibrium constant = 2.75 × 10⁻⁵

Now put all the given values in the above formula, we get:

`\Delta G=-40.82kJ/mol+(8.314* 10^(-3)kJ/mole.K)* (298K)* \ln (2.75* 10^(-5))`

`\Delta G=-66.8kJ/mol`

Therefore, the value of
`\Delta G` is, -66.8 kJ/mol