At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L reaction vessel, what will the equilibrium concentration of N2 be?

At equilibrium, the concentration of is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

+ ⇄ 2

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor   present. Immediately, and are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: []=0   ;     [ ]= 0    ; []=0.60M

C: []=+x   ;     [ ]= +x    ; []=-2x

E: []=0+x   ;     [ ]= 0+x   ; []=0.60-2x

Now we can use the constant information:

=

=

=

At equilibrium, the concentration of is going to be 0.30M

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.