The equilibrium constant at 850 K for the reaction N2(g) + O2(g) 2NO(g) is 0.5625. The equilibrium concentration of NO(g) is 3.0 × 10-3 M. If the equilibrium concentrations of N2(g) and O2(g) are equal, the concentration of N2(g) in M is

Question

The equilibrium constant at 850 K for the reaction N2(g) + O2(g) <=> 2NO(g) is 0.5625. The equilibrium concentration of NO(g) is 3.0 × 10-3 M. If the equilibrium concentrations of N2(g) and O2(g) are equal, the concentration of N2(g) in M is (A)  4.0 × 10-3 (B)  4.0 × 10-2 (C)  1.6 × 10-3 (D)  3.0 × 10-3

Tardigrade Admin · Accepted Answer

Given, N 2(g)+ O 2(g) leftharpoons 2 NO (g) i.e. equilibrium constant, KC=0.5625 Equilibrium concentration of [ NO ]=3 × 10-3 M ∵ KC=([ NO ]2/[ N 2][ O 2]) But, [ N 2]=[ O 2] (Given) ∴ KC=([ NO ]2/[ N 2]2) or, 0.5625=([3 × 10-3]2/[ N 2]2) or, [ N 2]2=([3 × 10-3]2/0.5625) =(9 × 10-6/0.5625) Or, [ N 2]2=16 × 10-6 or, [ N 2]=4 × 10-3

$4.0 \times 1 0^{- 3}$

$4.0 \times 1 0^{- 2}$

$1.6 \times 1 0^{- 3}$

$3.0 \times 1 0^{- 3}$

Q. The equilibrium constant at 850K for the reaction NX2(g)​​+OX2(g)​​​2NOX(g)​​ is 0.5625. The equilibrium concentration of NOX(g)​​ is 3.0×10−3M. If the equilibrium concentrations of NX2(g)​​ and OX2(g)​​ are equal, the concentration of NX2(g)​​ in M is

4.0×10−3

4.0×10−2

1.6×10−3

3.0×10−3

$4.0 \times 1 0^{- 3}$

$4.0 \times 1 0^{- 2}$

$1.6 \times 1 0^{- 3}$

$3.0 \times 1 0^{- 3}$