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vampirchik [111]
3 years ago
6

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 N

2O. Calculate K at this temperature for the following reaction. 2 N2(g) + O2(g) equilibrium reaction arrow 2 N2O(g)
Chemistry
1 answer:
zhenek [66]3 years ago
7 0

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

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