Question

1) The equilibrium constant (Keq) for the following reaction is 6.2x10-6. For the following assume a temperature of 25 °C.L-Malate + NAD+«oxaloacetate + NADH + H+Inthe cell the DG for the reaction is –10 kJ/mol and the ratio of [NADH]/[NAD] is 0.1. What must be the ratio of [oxaloacetate] to [L-malate]?

Answer 1

Answer:

The ratio of $\frac{Oxalaoacetate}{L-Malate} \,\, is\,\, 1.1\times10^{-6}$

Explanation:

The given chemical reaction is as follows

$L-MALATE + NAD^{+}\rightleftharpoons Oxalaoacetate+NADH+H^{+}$

$\bigtriangleup G^{o}\,=\,-RTlnK\\=-8314\times 298ln\times 6.2\times 10^{-6}\\\=-2477.572[ln6.2+ln10^{-6}]\\=-5705.84[log6.2+log10^{-6}]\\=29729.77$

$\bigtriangleup G\,= \bigtriangleup  G^{o}$

$-10=29.713+8.314\times298ln\frac{[Oxaloacetate][NADH]}{[L-Malate][NAD^{+}]}$

$\frac{-39.713}{2.477}=2.303[log\frac{[oxaloacetate]}{[L-malate]}+log\frac{[NADH]}{NAD}]$

$-6.96=log\frac{Oxalaoacetae}{L-Malate}+log10^{-1}$

$-5.96=log\frac{Oxalaoacetate}{L-Malate}$

$\frac{[Oxalaoacetate]}{[L-Malate]}=1.1\times10^{-6}$