At equilibrium, the concentrations of N2 = 3.0 x 10- 3 M, O2 = 4.2 × 10- 3 M and NO = 2.8 × 10- 3 M in a sealed vessel at 800 K. What will be Kc for the reaction, N2(g) + O2(g) leftharpoons 2NO(g) ?

Question

At equilibrium, the concentrations of N2 = 3.0 x 10- 3 M, O2 = 4.2 × 10- 3 M and NO = 2.8 × 10- 3 M in a sealed vessel at 800 K. What will be Kc for the reaction, N2(g) + O2(g) leftharpoons 2NO(g) ? (A)  522 (B)  6.2 (C)  0.622 (D)  8.5

Tardigrade Admin · Accepted Answer

For the reaction, equilibrium constant Kc can be written as, kc =([NO]2/[N2][O2]) =((2.8×10-3 M2)/(3.0 ×10-3 M)(4.2 ×10-3 M)) =0.622

Q. At equilibrium, the concentrations of $N_{2} = 3.0 x 1 0^{- 3} M, O_{2} = 4.2 \times 1 0^{- 3} M$ and $NO = 2.8 \times 1 0^{- 3} M$ in a sealed vessel at $800 K .$ What will be Kc for the reaction, $N_{2} (g) + O_{2} (g) ⇌ 2 N O_{(g)}$ ?

$522$

$6.2$

$0.622$

$8.5$

For the reaction, equilibrium constant $K_{c}$ can be written as,
$k_{c} = \frac{[ NO ] ^{2}}{[ N _{2} ] [ O _{2} ]}$
$= \frac{( 2.8 \times 1 0 ^{- 3} M ^{2} )}{( 3.0 \times 1 0 ^{- 3} M ) ( 4.2 \times 1 0 ^{- 3} M )} = 0.622$

Q. At equilibrium, the concentrations of N2​=3.0x10−3M,O2​=4.2×10−3M and NO=2.8×10−3M in a sealed vessel at 800K. What will be Kc for the reaction, N2​(g)+O2​(g)⇌2NO(g)​ ?

522

6.2

0.622

8.5