At equilibrium, the concentration of N _2=3.0 10^ -3 M , O _2=4.2… - Vidyadip

MCQ

At equilibrium, the concentration of $N _2=3.0 \times 10^{-3}$ $M , O _2=4.2 \times 10^{-3} M$ and $NO =2.8 \times 10^{-3} M$ in a sealed vessel at 800 K . Value of Kc will be :
$
N_2(g)+O_2(g) \rightleftharpoons 2 NO(g)
$

A
1.732
✓
0.622
C
0.335
D
0.104

✓

Answer

Correct option: B.

0.622

(B)0.622
$\begin{array}{l} K _{ c }=\frac{[ NO ]^2}{\left[N_2\right] \times\left[ O _2\right]} \\ \frac{\left(2.8 \times 10^{-3}\right)^2}{\left(3.0 \times 10^{-3}\right) \times\left(4.2 \times 10^{-3}\right)} \\ =0.622\end{array}$

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