Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous as written under cellular conditions?

Question

asked Feb 13, 2021 65.0k views

1 Answer

Alexmuller · Answer 1 · 2021-02-18T14:21:27+0000

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

$A->B;\Delta _rG^o=-13 kJ/mol$

$C ->D ;\Delta _rG^o=3.5 kJ/mol$

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

$Q_1=(4.0mM)/(0.050mM)=80\\ Q_2=(0.010mM)/(0.060mM)=0.167$

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

$K_1=exp(-(-13000J/mol)/(8.314J/mol*K*303.15K) )=173.8\\K_2=exp(-(3500J/mol)/(8.314J/mol*K*303.15K) )=0.249$

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.