# 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?

Dec 28, 2014

The answer is $3.3 M$.

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 $N {O}_{2}$, $A$, was placed in the flask, which means that the initial concentrations of $N O$ and ${O}_{2}$ were zero.

We must use the reverse reaction, since no amounts of $N O$ and ${O}_{2}$ are present at the start; this means that the equlibrium constant will be equal to

${K}_{r e v e r s e} = \frac{1}{K} _ \left(f o r w a r d\right) = \frac{1}{24.0} = 0.0417$

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 $N {O}_{2}$ at equlibrium (and of course, before the reaction) must be greater than those of $N O$ and ${O}_{2}$.

.....$2 N {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 N {O}_{\left(g\right)} + {O}_{2 \left(g\right)}$
I.........A.................0...................0
C......-2x...............+2x................+x
E.......(A-2x)...........2x.................x

Since we know that $N O$'s equilibrium concentration is 0.800 M, we get $2 x = 0.800 \to x = \frac{0.800}{2} = 0.400 M$

The expression for the equilibrium constant is

${K}_{e q} = \frac{{\left[N O\right]}^{2} \cdot \left[{O}_{2}\right]}{{\left[N {O}_{2}\right]}^{2}} = \frac{{0.800}^{2} \cdot 0.400}{A - 0.800} ^ 2 = 0.0417$

${\left(A - 0.800\right)}^{2} = \frac{{0.800}^{2} \cdot 0.400}{0.0417} = 6.14$

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, $N {O}_{2}$'s initial concentration is $3.3 M$.

Notice that the prediction about $N {O}_{2}$'s concentration at equilibrium was accurate; at equilibrium, the species have the following concentrations:

$\left[N {O}_{2}\right] = 3.3 - 0.8 = 2.5 M$
$\left[N O\right] = 0.8 M$
$\left[{O}_{2}\right] = 0.4 M$

Dec 28, 2014

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.

$2 \text{NO" + "O"_2 ⇌ 2"NO"_2}$

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

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

So $x = 0.400$

Then

["NO"]_"eq" = 0.800

${\left[{O}_{2}\right]}_{\text{eq}} = x = 0.400$

[NO_2]_"eq" = C – 2x = C -0.800

Step 3. Write the ${K}_{\text{eq}}$ expression.

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

Step 4. Insert these values into the ${K}_{\text{eq}}$ 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

${\left[{O}_{2}\right]}_{\text{eq}} = x = 0.400$

[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