Question

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and 0.040 mol/L of O2, calculate the equilibrium concentration of NO in mol/L

how would this be done?

Answer 1

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x.  So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

Answer 2

We have that The equilibrium concentration of NO in mol/L is

$[NO]=0.011mol/L$

From the question we are told that:

Equation:

$N2(g) + O2(g) ⇄ 2NO(g)$

$Kc = 0.10 at 2000 \textdegree C$

Concentration of $N_2$ $C_n=0.040 mol/L$

Concentration of $O_2$ $C_n=0.040 mol/L$

Generally, the equation for KC this is mathematically given by

$KC=\frac{(NO)^2}{[N_2][O_2]}$

$[NO]^2=0.10*0.040*0.040$

$[NO]^2=1.6*10^{-4}$

$[NO]=0.011mol/L$

 

In conclusion

The equilibrium concentration of NO in mol/L is

$[NO]=0.011mol/L$

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