When one mole of A and one mole of B were heated in a one litre flask at T ( K ), 0.5 moles of C was formed in the equilibrium: A + B leftharpoons C + D The equilibrium constant Kc is.

Question

When one mole of A and one mole of B were heated in a one litre flask at T ( K ), 0.5 moles of C was formed in the equilibrium: A + B leftharpoons C + D The equilibrium constant Kc is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/42c27098fa43f684a4de3f69d0b7325f-.png" /> ∴ Kc=([C][D]/[A][B])=(0.5 × 0.5/0.5 × 0.5)=1

Q. When one mole of $A$ and one mole of $B$ were heated in a one litre flask at $T (K), 0.5$ moles of $C$ was formed in the equilibrium:
$A + B ⇌ C + D$
The equilibrium constant $K_{c}$ is_____.

$∴ K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{0.5 \times 0.5}{0.5 \times 0.5} = 1$

Q. When one mole of A and one mole of B were heated in a one litre flask at T(K),0.5 moles of C was formed in the equilibrium: A+B⇌C+D The equilibrium constant Kc​ is_____.

∴Kc​=[A][B][C][D]​=0.5×0.50.5×0.5​=1

Q. When one mole of $A$ and one mole of $B$ were heated in a one litre flask at $T (K), 0.5$ moles of $C$ was formed in the equilibrium:
$A + B ⇌ C + D$
The equilibrium constant $K_{c}$ is_____.

$∴ K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{0.5 \times 0.5}{0.5 \times 0.5} = 1$