Homework Help 11.59

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John Huang 1G
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Homework Help 11.59

Postby John Huang 1G » Sun Nov 26, 2017 10:10 am

For problem 11.59, I understand that we would be using an ICE chart. But what steps would I take next to find the equilibrium concentrations of each substance?

AtreyiMitra2L
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Re: Homework Help 11.59

Postby AtreyiMitra2L » Sun Nov 26, 2017 11:45 am

First, you would have to change moles to molar concentrations. Then you would put all your initial molar concentrations in your ice chart. Then for change, based on the coefficients of the reactants and products, subtract/add the coefficient x x. You subtract for the reactants and add for the products. Then get out you equilibrium constant expression with products on top and reactants on the bottom. Given that you know the equilibrium constant, you can solve for x with the quadratic equation.

Allyson Charco Dis1G
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Re: Homework Help 11.59

Postby Allyson Charco Dis1G » Sun Nov 26, 2017 6:49 pm

A reaction mixture is prepared by mixing 0.100 mol SO2, 0.200 mol NO2, 0.100 mol NO, and 0.150 mol SO3 in
a 5.00-L reaction vessel. The reaction SO2(g) + NO2(g) ∆ NO(g) + SO3(g) is allowed to reach equilibrium at 460 C, when Kc 85.0. What is the equilibrium concentration of each substance?
So first we would have to calculate the initial concentrations of each of the species. O2(g) + NO2(g) ∆ NO(g) + SO3(g)
Which are [SO2]= 0.100 mol/5.00 L which equals 0.0200 molL^-1
[NO]= 0.100 mol/5.00L which equals 0.0200molL^-1
[NO2]= 0.200mol/5.00L equals 0.0400molL^-1
[SO3]= 0.150/5.00L = 0.0300molL^-1
We use these to calculate Q
Q= (0.0200)(0.0300)/(0.0200)(0.0400)=0.75
Because 0.75 is less one 1.
Kc =[NO][SO3]/[SO2][NO2] which equals 85.0= (0.0200 + x) (0.0300 +x)/(0.0200 -x)(0.0400 -x) so you foil it out
which expands to (x^2+0.500x+0.000600)/(x^2-0.0600x + 0.000800)
Then you multiply 85.0 times (x^2 - 0.0600x +0.000800) = (x^2+0.500x+0.000600)
84.0X^2-5.15X+0.0674=0
then you simply and have x= -(-5.15) pus or minus square root of (5.150)^2-4(84.0)(0.0674)/2(84.0)
X= +0.0424 OR +0.0189
IN WHICH =0,0424 is not used because it is larger than the 0.400molL^-1 of NO2
then we would proceed to use the final value that we got before to find the equilibrium constant
[SO2]= 0.0200molL^-1 - 0.0189molL-1 =0.0011molL^-1
[NO2]=0.400molL^-1 - 0.0189molL-1 = 0.0211molL^-1
[NO]=0.0200molL^-1 + 0.0189molL-1 = 0.0389molL^-1
[SO3]= 0.0300 molL^-1 +0.0189molL-1 = 0.0489molL-1 then we would plug it in The Kc equation with the values we just got
Kc= (0.0389)(0.0489)/(0.0011)(0.0211)= 82.0
which we can determine the 85 and 82 are relatively close values given to all the limitations and the nature of the calculation.


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