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?

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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?

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Answer 1 · 2014-12-28T16:59:39

The answer is $3.3 M$ $3.3M$ .

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 $N O_{2}$ $NO_2$ , $A$ $A$ , was placed in the flask, which means that the initial concentrations of $N O$ $NO$ and $O_{2}$ $O_2$ were zero.

We must use the reverse reaction, since no amounts of $N O$ $NO$ and $O_{2}$ $O_2$ are present at the start; this means that the equlibrium constant will be equal to

$K_{r e v e r s e} = \frac{1}{K_{f o r w a r d}} = \frac{1}{24.0} = 0.0417$ $K_(reverse) = 1/K_(fo rward) = 1/24.0 = 0.0417$

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 $N O_{2}$ $NO_2$ at equlibrium (and of course, before the reaction) must be greater than those of $N O$ $NO$ and $O_{2}$ $O_2$ .

..... $2 N O_{2 (g)} ⇌ 2 N O_{(g)} + O_{2 (g)}$ $2NO_(2(g)) rightleftharpoons 2NO_((g)) + O_(2(g))$
I.........A.................0...................0
C......-2x...............+2x................+x
E.......(A-2x)...........2x.................x

Since we know that $N O$ $NO$ 's equilibrium concentration is 0.800 M, we get $2 x = 0.800 \to x = \frac{0.800}{2} = 0.400 M$ $2x = 0.800 -> x= 0.800/2 = 0.400M$

The expression for the equilibrium constant is

$K_{e q} = \frac{{[N O]}^{2} \cdot [O_{2}]}{{[N O_{2}]}_{2}^{2}} = \frac{{0.800}^{2} \cdot 0.400}{{(A - 0.800)}^{2}} = 0.0417$ $K_(eq) = ([NO]^2 * [O_2])/([NO_2]^2) = (0.800^2 * 0.400)/(A-0.800)^2 = 0.0417$

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

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

Therefore, $N O_{2}$ $NO_2$ 's initial concentration is $3.3 M$ $3.3M$ .

Notice that the prediction about $N O_{2}$ $NO_2$ 's concentration at equilibrium was accurate; at equilibrium, the species have the following concentrations:

$[N O_{2}] = 3.3 - 0.8 = 2.5 M$ $[NO_2] = 3.3 - 0.8 = 2.5M$
$[N O] = 0.8 M$ $[NO] = 0.8M$
$[O_{2}] = 0.4 M$ $[O_2] = 0.4M$

Answer 2 · 2014-12-28T17:58:16

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 $C$ $C$ = the initial concentration of NO.

$2 NO + O_{2} ⇌ 2 {NO}_{2}$ $2"NO" + "O"_2 ⇌ 2"NO"_2"$

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

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

So $x = 0.400$ $x = 0.400$

Then

${[NO]}_{eq} = 0.800$ $["NO"]_"eq" = 0.800$

${[O_{2}]}_{eq} = x = 0.400$ $[O_2]_"eq" = x = 0.400$

$[NO_2]_"eq" = C – 2x = C -0.800$

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:

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

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

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

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

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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?

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