# A student places 0.025 moles of solid sulfur and 0.100 moles of nitrogen monoxide gas in a closed 1.00L container and 0°c for several days. Calculate the number of molecules of NO in the 1.00L container after equilibrium is reached, given the following thermodynamic data.

I have calculated ∆G and ∆S already and set up an ICF box but i'm having trouble setting up the ICE box to find the molecules of NO.

S + 4NO --> SO2 + 2N2O

I 0.025 0.100 0 0

C 0.025 0.100 +0.25 +2x+0.25

F 0 0 0.25x (o.25+2x)^2

I know to solve for the moles and molecules of NO I would need to do the reverse reaction but not sure how to set that up.

## You didn't provide any of the data but my first impression is that you do not need to calculate anything in reverse. If you have delta G, then use dG = -RT*ln*K (solve for K) and set up the ICE box as

........S + 4NO ==> SO2 + 2N2O

I...0.025..0.1.......0.....0

C.....-x...-4x.......x.....2x

E.....??..0.1-4x.....x.....2x

Plug the E line into the Keq expression and solve for x and 0.1-4x. Important note: the SOLID S is NOT included in the K expression; i.e.,

Keq = (N2O)(SO2)/(NO)^4

After you have 0.1-4x convert that to mols and to molecules.

## The thermodynamic data:

s No So2 N2O

S 31.9 211 256 220

∆H 0 90.4 -297 81.6

I solve for ∆G and got an answer of -446287.3 then I used that to find K which is 2.48 x 10^85.

I set up the ice box just like you showed but do I set it equal to K?

(2x)(X)/0.1-4x= 2.48 x10^85 or do I set it equal to ∆G. I'm not sure if that even makes sense, and then after having 0.1-4x and finding x, what would be the units of that to convert it to moles and molecules?

## If K = 2.48E85, then note you didn't square or to the 4th power where you should and the complete equation would be

2.48E85 = (2x)^2(x)/(0.1-4x)^4

I see I omitted the square term for (NO)^2 in my first response.

The units of x are mols/L = M so since you have 1.0L that essentially gives moles directly and 1 mol contains 6.022E23 molecules.

Having said all of that, I am reconsidering my first response. With a huge K such as this AND the fact they give you (NO) at the beginning, I believe I would first consider the reaction going far to completion (I didn't realize earlier just how large that K was AND that the NO concn was given) and doing the reverse is about the only way to get at it. Otherwise, plugging those values in the above equation will give you essentially 0.025 for (SO2) and zero for (NO). It will be very very small BUT it won't be zero and the only way to get it I think is to do the reverse equation as you first suggested. If you post again, please repost ALL of this at the top of the page. It's on page 3 now and I don't look much past page 2. I looked specifically for this one to see if there were follow up questions and that's how I found it. Hope this helps. By the way, you haven't done any work needlessly; i.e., you can use everything you have done so far except for solving that Keq part which you didn't do anyway. :-). Here is the reverse set up but check it out closely.

..........S + 4NO ==> SO2 + 2N2O

I.........0....0......0.025 + 0.050

C.........x...4x......-x.....-2x

E........solid.4x..0.025-x..0.050-2x

K = (SO2)(N2O)^2/(NO)^4

K = (0.025-x)(0.050-2x)^2/((4x)^4

and solve for x

## To set up the ICE box for the reverse reaction, you need to identify the stoichiometric coefficients. In the given chemical equation,

S + 4NO --> SO2 + 2N2O,

the stoichiometric coefficient of NO is 4. Therefore, the reverse reaction would be:

SO2 + 2N2O --> S + 4NO.

Now, let's set up the ICE box for this reverse reaction:

SO2 + 2N2O --> S + 4NO

I 0 0 0.025 0.100

C +0.25x +x -0.025 -0.100

E 0.025+0.25x 0.100+x 0 0

Note that the change for S and NO is negative because they are being consumed in the reaction. The change for SO2 and N2O is positive because they are being formed.

To solve for the equilibrium concentrations, you would plug the values from the "E" row into the equilibrium expression, which would be an equation based on the reverse reaction:

K = [NO]^4 / ([S][SO2][N2O]^2)

Since you have already calculated ΔG and ΔS, you can use them to find the equilibrium constant (K) using the equation:

ΔG = -RT ln(K),

where R is the ideal gas constant and T is the temperature in Kelvin.

Once you have the equilibrium constant (K), you can plug it into the equilibrium expression and solve for [NO]. Multiply the calculated value of [NO] by Avogadro's number to get the number of molecules of NO in the 1.00L container after equilibrium is reached.