Given N2(g)+O2(g)<--> 2NO(g) Keq=8.00 *10-2 at 500 C
In a 2.00L flask, 0.4 mol of N2 and 0.4 mol of oxygen gas, O2, reacted to equilibrium. Establish the equilibrium concentration of all the gases.
(Answer: [N2]=0.175 mol/L; [O2] = 0.175 mol/L; [NO] = 0.0496 mol/L) This one has got a quadratic I can't figure out
My first step: $\frac{\left(2x\right)}{\left(0.2-x\right)^2}=0.08$(2x)(0.2−x)2 =0.08
My second step: $0.08\cdot\left(.2-x\right)\left(.2-x\right)=2x$0.08·(.2−x)(.2−x)=2x
Don't know what's next
8 years ago
Answered By Charles S
Yes, you set it up perfectly. Just a little math from here.
$0.08\left(.04-.4x+x^2\right)=2x$0.08(.04−.4x+x2)=2x Use the distributive property here
$0.032-0.32x+0.08x^2=2x$0.032−0.32x+0.08x2=2x Again... use the distributive property
$0.08x^2-2.32x-0.032=0$0.08x2−2.32x−0.032=0 bring to one side and get into standard form
and from here... a = 0.08, b = -2.32, c = -0.032
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a
Once you have x, you can substitute into the ice table to find the equilibrium values for each of the compounds. Also note -- At the time of this writing, chem 30 in Alberta has removed questions involving theh quadratic formula. See the whiteboard below for an explanation of why you cannot approximate and cancel x out.
Attached Whiteboard:
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