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?

Respuesta :

the answer is 104.9, i dont know how though

The chemical reaction is given as:

[tex]2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}[/tex]

Here, oxygen is oxidised and [tex]NaDP^{+}[/tex] is reduced. Thus, redox reaction occurs.

For cell reaction, [tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex]          (2)

where, [tex]\Delta G^{o} [/tex] = standard state free energy

n= number of electrons

F= Faraday constant ([tex]96485.33 C/mol[/tex])

[tex]E^{o}_{cell}[/tex] = 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  [tex]\Delta G^{o} [/tex].

Now, calculate the value of cell potential

[tex]E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}[/tex]           (1)

[tex]E^{o}_{cathode}[/tex] = [tex]-0.324 V[/tex] (standard reduction potential of [tex]NaDP^{+}[/tex])

[tex]E^{o}_{anode}[/tex] = [tex]1.23 V[/tex] (standard reduction potential of [tex]O_{2}[/tex])

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

[tex]E^{o}_{cell} = -0.324 V-1.23 V[/tex]

= [tex]-1.554 V[/tex]

Now, substitute above value in formula (2)

[tex]\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V) [/tex]  

= [tex]299876.40564 CV/mol[/tex]

Since, one coulomb volt is equal to one joule.

Thus, value of [tex]\Delta G^{o}[/tex] is equal to [tex]299876.40564 J/mol[/tex] or [tex]299.87640564 kJ/mol[/tex]