1 mole of NO and 1 mole of O3 are taken in a 10 L vessel and heated. At equilibrium, 50 % of NO (by mass) reacts with O3 according to the equation: NO(g) +O3(g) leftharpoons NO2(g) +O2(g). What will be the equilibrium constant for this reaction?

Question

1 mole of NO and 1 mole of O3 are taken in a 10 L vessel and heated. At equilibrium, 50 % of NO (by mass) reacts with O3 according to the equation: NO(g) +O3(g) leftharpoons NO2(g) +O2(g). What will be the equilibrium constant for this reaction? (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ccb8ecfa037999c82fd208a7aa67d75e-.png" /> Kc=(0.05×0.05/0.05×0.05)=1

Q. $1$ mole of $NO$ and $1$ mole of $O_{3}$ are taken in a $10 L$ vessel and heated. At equilibrium, $50%$ of $NO$ (by mass) reacts with $O_{3}$ according to the equation:
$N O_{(g)} + O_{3 (g)} ⇌ N O_{2 (g)} + O_{2 (g)}$ .
What will be the equilibrium constant for this reaction?

$1$

$2$

$3$

$4$

$K_{c} = \frac{0.05 \times 0.05}{0.05 \times 0.05} = 1$

Q. 1 mole of NO and 1 mole of O3​ are taken in a 10L vessel and heated. At equilibrium, 50% of NO (by mass) reacts with O3​ according to the equation: NO(g)​+O3(g)​⇌NO2(g)​+O2(g)​. What will be the equilibrium constant for this reaction?

1

2

3

4

Kc​=0.05×0.050.05×0.05​=1

Q. $1$ mole of $NO$ and $1$ mole of $O_{3}$ are taken in a $10 L$ vessel and heated. At equilibrium, $50%$ of $NO$ (by mass) reacts with $O_{3}$ according to the equation:
$N O_{(g)} + O_{3 (g)} ⇌ N O_{2 (g)} + O_{2 (g)}$ .
What will be the equilibrium constant for this reaction?

$1$

$2$

$3$

$4$

$K_{c} = \frac{0.05 \times 0.05}{0.05 \times 0.05} = 1$