Gen chem, Gibbs free energy

[Impulse155]

The question states
Given that the standard enthalpy of formation for NO(g) is 90.25 kj/mole, calclate the free energy change for the following reactions at 25oC

N2(g) + O2(g) <--> 2NO(g)

I know the equation is /\G= /\H-T/\S
It gives the bond entropies and when you calculate the /\S you get 25.
The temperature is 25oC, convert to kelvin you get 298
The 90.25 is per mole, your producing 2 moles of NO(g) so it would be 90.25 x 2 = 180.5
So I would say its /\G = 180.5 - (298oC)(25)
but then there is a 1x10^-3 after that so the answer is
/\G = 180.5 - (298oC)(25)(1x10^-3)

Where did this 10^-3 come from?
This is Top Score gen chem test 1.

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[OUDent88]

make sure u pay attention to your units...enthalpies are given as kj/mole while entropy is given in J. The 10^-3 comes from converting the units so they are the same.

 

[Impulse155]

make sure u pay attention to your units...enthalpies are given as kj/mole while entropy is given in J. The 10^-3 comes from converting the units so they are the same.

Err I don't have it open but thats definitely what the problem was :|

 

[videlmegami]

Bringing alive a totally dead thread here...but I thought since I came and searched for this, there may be others as well. Anyways, OUDent is actually right here...in the table given, entropy was given in Joules...which means that to convert it into kJ (the unit in which enthalpy was given), we need to multiply the equation by 1*10^-3.

Therefore, the G=H-TS ends up being: G = 2(90.25kJ)-(273+25)(25J)(1*10^-3)

I originally was confused on how the entropy was positive, but then I realized that we had 2 moles of gas in the reactants and two moles in the products, thus, it's probable that the entropy did increase. This was nothing to do with OP's question...just something I was originally confused about and then figured out so maybe this may help future TopScore test takers...