Hey guys! Stuck on another one: The reaction 2N2O(g) 2N2(g) + O2(g) has Kc = 3.5 × 10-18 at a particular temperature. If 0.20 mol of N2O is placed in a 1.00 liter container, what will be the N2 concentration when equilibrium is reached?
1. 4.4 × 10-9 mol/liter 4. 3.3 × 10-7 mol/liter
2. 3.2 × 10-8 mol/liter 5. none of the previous answers
3. 4.5 × 10-6 mol/liter
My maybe answer is : 5 none of these
I did an ICE table
and got my Kc equation to be Kc-3.5*10^-18=(2x)^2)x/(.2-2x)^2, I got a cubic, and solved getting x=.207, when I plug that in I don't get any of the above answers. I am pretty positive that I did this question completely wrong, so can someone please help? Thanks so much!!
2 answers
Show your work and we can check it.
Sorry! Here:
ICE Table
N2O N2 O2
.2M 0M 0M
-2x 2x x
.2-2x 2x x
Kc=[N2]^2[O2]/[N2O]^2
3.5*10^-18=[2x]^2[x]/[.2-2x]^2
0=-4x^3+1.4*10^-17x^2 -2.8*10^-18+1.4*10^-19
I then graphed this cubic and my x was actually 3.3*10^-7
I plugged that x into 2x to get 6.6*10^-7, which is none of the above
ICE Table
N2O N2 O2
.2M 0M 0M
-2x 2x x
.2-2x 2x x
Kc=[N2]^2[O2]/[N2O]^2
3.5*10^-18=[2x]^2[x]/[.2-2x]^2
0=-4x^3+1.4*10^-17x^2 -2.8*10^-18+1.4*10^-19
I then graphed this cubic and my x was actually 3.3*10^-7
I plugged that x into 2x to get 6.6*10^-7, which is none of the above
2 answers