Question

An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)--> 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?

Answer 1

The answer is `3.3M`.

The best way to approach this problem is by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart). We know that, initially, a concentration of `NO_2`, `A`, was placed in the flask, which means that the initial concentrations of `NO` and `O_2` were zero.

We must use the reverse reaction, since no amounts of `NO` and `O_2` are present at the start; this means that the equlibrium constant will be equal to

`K_(reverse) = 1/K_(fo rward) = 1/24.0 = 0.0417`

SInce the equilibrium constant for the reverse reaction is smaller than 1, the final mixture will contain mostly reactants, which means that the concentration of `NO_2` at equlibrium (and of course, before the reaction) must be greater than those of `NO` and `O_2`.

.....`2NO_(2(g)) rightleftharpoons 2NO_((g)) + O_(2(g))`
I.........A.................0...................0
C......-2x...............+2x................+x
E.......(A-2x)...........2x.................x

Since we know that `NO`'s equilibrium concentration is 0.800 M, we get `2x = 0.800 -> x= 0.800/2 = 0.400M`

The expression for the equilibrium constant is

`K_(eq) = ([NO]^2 * [O_2])/([NO_2]^2) = (0.800^2 * 0.400)/(A-0.800)^2 = 0.0417`

`(A-0.800)^2 = (0.800^2 * 0.400)/0.0417 = 6.14`

Solving for `A` will produce two values, one negative and one positive; since `A` represents concentration, which cannot be negative, the only value for `A` will be 3.3.

Therefore, `NO_2`'s initial concentration is `3.3M`.

Notice that the prediction about `NO_2`'s concentration at equilibrium was accurate; at equilibrium, the species have the following concentrations:

`[NO_2] = 3.3 - 0.8 = 2.5M`
`[NO] = 0.8M`
`[O_2] = 0.4M`

Answer 2

The initial concentration of the NO₂ was 3.279 mol/L.

Step 1. Write the balanced chemical equation for the equilibrium.

2NO + O₂ ⇌ 2NO₂

Step 2. Set up an ICE table:

Let `C` = the initial concentration of NO.

`2"NO" + "O"_2 ⇌ 2"NO"_2"`

I/mol·L⁻¹: 0; 0; `C`
C/ mol·L⁻¹: +2`x`; +`x`; -`2x`
E/ mol·L⁻¹: 2`x`; `x`; `C - 2x`

We know that `["NO"]_"eq" = 2x = 0.800`

So `x = 0.400`

Then

`["NO"]_"eq" = 0.800`

`[O_2]_"eq" = x = 0.400`

`[NO_2]_"eq" = C – 2x = C -0.800`

Step 3. Write the `K_"eq"` expression.

`K_"eq" = ["NO"_2]^2/(["NO"^2]["O"_2]) = 24.0`

Step 4. Insert these values into the `K_"eq"` expression.

`K_"eq" = ["NO"_2]^2/(["NO"^2]["O"_2])= (C - 0.800)^2/(0.800^2 × 0.400) = 24.0`

Step 5. Solve for `C`.

`(C-0.800)^2 = 24.0 × 0.640 × 0.400 = (0.800^2 × 0.400) = 6.144`

`C^2 – 1.600C + 0.640 = 6.144`

`C^2 – 1.600C – 5.504 = 0`

`C = 3.279`

The initial concentration of the NO₂ was 3.279 mol/L.

Check:

`["NO"]_"eq" = 0.800`

`[O_2]_"eq" = x = 0.400`

`[NO_2]_"eq" = C -2x = 3.279 – 0.800 = 2.479`

`K_"eq" = ["NO"_2]^2/(["NO"^2]["O"_2]) = 2.479^2/(0.800^2 × 0.400) = 24.0`