Question

Calculate ΔG°' for the reaction A + B <-> C + D at 25°C when the equilibrium concentrations are [A] = 10 uM, [B] = 15 uM, [C] = 3 uM, and [D] = 5 uM. R = 8.314 J/mol⋅K. Give your answer to the nearest hundredths in J/mol, but only provide the numerical value (i.e. don't include units in your answer).

Answer 1

The value of ΔG° of the reaction is 5704.82 J/mol.

Step-by-step explanation:

A + B ⇄ C + D

Equilibrium concentrations of reactants an products:

[A] = 10 μM, [B] = 15 μM, [C] = 3 μM, and [D] = 5 μM

An equilibrium constant of the reaction can be written as:

`K=([C][D])/([A][B])`

`K=(3 \mu M* 5 \mu M)/(10 \mu M* 15 \mu M)=0.1`

`\Delta G^o=-RT\ln K_1`

where,

R = Gas constant =
`8.314J/K mol`

T = temperature =
`25^oC=[273+25]K=298K`

`K_1` = equilibrium constant at 25°C = 0.1

Putting values in above equation, we get:

`\Delta G^o=-(8.314J/Kmol)* 298K* \ln (0.1)`

`\Delta G^o=5704.82 J/mol`

The value of ΔG° of the reaction is 5704.82 J/mol.