**Find the free energy at 25 degrees for the nonstandard conditions at point C where the reaction?**

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*Calculate the free energy delta G at 25 degrees Celsius for the nonstandard conditions at point C where the reaction quotient is 3.58*10^9*

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*I know you use the equation,

delta G = delta G (degrees sign) + RTln(Q)

The delta G (degrees sign) is -40.28

I got answers like -54.4,-14.22,-94.68,94.68,14.22 kJ and it's still wrong. It makes no sense.

Be sure to put delta G in joules.

delta G = delta Go + RTlnQ = -40,280 J + (8.31 J/mole K)(298 K) ln (3.58 x 10^9)

= -40,280 J + 54,480 J = 14,200 J = +14.2 kJ

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calculate the free energy at 25 for the nonstandard conditions at point c where the reaction quotient is