Question

What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?

Answer 1

the answer is 104.9, i dont know how though

Answer 2

The chemical reaction is given as:

`2NaDP^(+) +2H_(2)O\rightarrow 2NaDPH+2H^(+)+O_(2)`

Here, oxygen is oxidised and
`NaDP^(+)` is reduced. Thus, redox reaction occurs.

For cell reaction,
`\Delta G^(o) = -nFE^(o)_(cell)` (2)

where,
`\Delta G^(o)` = standard state free energy

n= number of electrons

F= Faraday constant (
`96485.33 C/mol`)

`E^(o)_(cell)` = cell potential

Substitute the value of number of electrons i.e. 2, Faraday constant and cell potential in the formula to determine the value of
`\Delta G^(o)`.

Now, calculate the value of cell potential

`E^(o)_(cell) = E^(o)_(cathode)- E^(o)_(anode)` (1)

`E^(o)_(cathode)` =
`-0.324 V` (standard reduction potential of
`NaDP^(+)`)

`E^(o)_(anode)` =
`1.23 V` (standard reduction potential of
`O_(2)`)

Put the above values in formula (1), we get:

`E^(o)_(cell) = -0.324 V-1.23 V`

=
`-1.554 V`

Now, substitute above value in formula (2)

`\Delta G^(o) = -2* 96485.33 C/mol *(-1.554 V)`

=
`299876.40564 CV/mol`

Since, one coulomb volt is equal to one joule.

Thus, value of
`\Delta G^(o)` is equal to
`299876.40564 J/mol` or
`299.87640564 kJ/mol`