Question

Given the following: [G3P] = 1.5x10-5M; [BPG] = 3.0x10-3M ; [NAD+] = 1.2x10-5M; [NADH]=1.0x10-4 ; [HPO42-]= 1.2x10-5 M; pH = 7.5 ; DGo=6.3 kJ/mol

Glyceraldehyde3-phosphate + NAD+ + HPO42- ---> 1,3-Biphosphoglycerate + NADH + H+

Predict whether this reaction will be spontaneous.

Answer 1

Answer: The given reaction is non-spontaneous in nature.

Explanation:

To calculate the $H^+$ concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

pH of the solution = 7.5

$7.5=-\log [H^+]$

$[H^+]=10^{-7.5)=3.1\times 10^{-8}M$

For the given chemical equation:

$\text{Glyceraldehyde3-phosphate }+NAD^++HPO_4^{2-}\rightarrow \text{1,3-Biphosphoglycerate }+NADH+H^+$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 6.3 kJ/mol = 6300 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = Temperature = $25^oC=[273+25]K=298K$  

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{[BPG][NaDH][H^+]}{[G_3P][NAD^+][HPO_4^{2-}]}$

$[BPG]=3.0\times 10^{-3}M$

$[NADH]=1.0\times 10^{-4}M$

$[H^+]=3.1\times 10^{-8}M$

$[G_3P]=1.5\times 10^{-5}M$

$[NAD^+]=1.2\times 10^{-5}M$

$[HPO_4^{2-}]=1.2\times 10^{-5}M$

Putting values in above equation, we get:

$\Delta G=6300J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(3.0\times 10^{-3})\times (1.0\times 10^{-4})\times (3.1\times 10^{-8})}{(1.5\times 10^{-5})\times (1.2\times 10^{-5})\times (1.2\times 10^{-5})}))\\\\\Delta G=9917.02J/mol=9.92kJ/mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

As, the Gibbs free energy of the reaction is positive. The reaction is said to be non-spontaneous.

Hence, the given reaction is non-spontaneous in nature.