Question

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N2O. Calculate K at this temperature for the following reaction. 2 N2(g) + O2(g) equilibrium reaction arrow 2 N2O(g)

Answer 1

Answer : The value of equilibrium constant (K) is, $5.71\times 10^4$

Explanation :

First we have to calculate the concentration of $N_2,O_2\text{ and }N_2O$

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M$

and,

$\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

The expression for equilibrium constant is:

$K=\frac{[N_2O]^2}{[N_2][O_2]}$

Now put all the given values in this expression, we get:

$K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}$

$K=5.71\times 10^4$

Thus, the value of equilibrium constant (K) is, $5.71\times 10^4$

Answer 2

Therefore, the equilibrium constant for the reaction is $4.08\times 10^{8}$.

Equilibrium Constant:

A chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.

Following is the equilibrium reaction occurring between $N_{2}$ and $O_{2}$.$2N_{2}\left ( g \right )+O_{2}\left ( g \right )\rightleftharpoons 2N_{2}O\left ( g \right )$

Expression for the equilibrium constant is as follows

$K=\frac{\left [ N_{2}O \right ]^{2}}{\left [ N_{2} \right ]^{2}\left [ O_{2} \right ]} ......(1)$

We have the following information.

The volume of the reaction flask$=2.00 \ L$

The concentration of $N_{2}\left ( g \right )$ at the equilibrium $=2.80\times10^{-4}mol$

The concentration of $O_{2}\left ( g \right )$ at the equilibrium $=2.50\times10^{-5}mol$

The concentration of $N_{2}O\left ( g \right )$ at the equilibrium $=2.00\times 10^{-2} mol$

Calculation of equilibrium concentration:

$\left [ N_{2} \left ( g \right )\right ]_{equ}=\frac{2.80\times 10^{-4}mol}{2.00L}$

$=1.40\times 10^{-4}M$

$\left [ O_{2}\left ( g \right ) \right ]_{equ}=\frac{2.50\times 10^{-5}mol}{2.00L}$

$=1.25\times 10^{-5}M$

$\left [ N_{2}O\left ( g \right ) \right ]_{equ}=\frac{2.00\times 10^{-2}mol}{2.00L}$

$=1.00\times 10^{-2}M$

Now substitute the concentrations in equilibrium constant expression $K=\frac{\left [ N_{2}O \right ]^{2}}{\left [ N_{2} \right ]^{2}\left [ O_{2} \right ]}\\K=\frac{\left ( 1.00\times 10^{-2} \right )^{2}}{\left ( 1.40\times 10\times -4 \right )^{2}\left ( 1.25\times 10^{-5} \right )}$

$=4.08\times10^{8}$

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