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

The ratio of
`(Oxalaoacetate)/(L-Malate) \,\, is\,\, 1.1*10^(-6)`

Step-by-step explanation:

The given chemical reaction is as follows

`L-MALATE + NAD^(+)\rightleftharpoons Oxalaoacetate+NADH+H^(+)`

`\bigtriangleup G^(o)\,=\,-RTlnK\\=-8314* 298ln* 6.2* 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*298ln([Oxaloacetate][NADH])/([L-Malate][NAD^(+)])`

`(-39.713)/(2.477)=2.303[log([oxaloacetate])/([L-malate])+log([NADH])/(NAD)]`

`-6.96=log(Oxalaoacetae)/(L-Malate)+log10^(-1)`

`-5.96=log(Oxalaoacetate)/(L-Malate)`

`([Oxalaoacetate])/([L-Malate])=1.1*10^(-6)`