The equilibrium constant for the reaction, H 2( g )+ I 2( g ) leftharpoons 2 HI ( g ) is 32 at a given temperature. The equilibrium concentrations of I 2 and HI are 0.5 × 10-3 and 8 × 10-3 M respectively. The equilibrium concentration of H 2 is

Question

The equilibrium constant for the reaction, H 2( g )+ I 2( g ) leftharpoons 2 HI ( g ) is 32 at a given temperature. The equilibrium concentrations of I 2 and HI are 0.5 × 10-3 and 8 × 10-3 M respectively. The equilibrium concentration of H 2 is (A) 1 × 10-3 M (B) 0.5 × 10-3 M (C) 2 × 10-3 M (D) 4 × 10-3 M

Tardigrade Admin · Accepted Answer

H 2+ I 2 leftharpoons 2 HI K=([ HI ]2/[ H 2][ I 2]) [ H 2]=([ HI ]2/[I2] × K)=((8 × 10-3)2/(0.5 × 10-3)(32)) =4 × 10-3 M

Q. The equilibrium constant for the reaction,
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
is $32$ at a given temperature. The equilibrium concentrations of $I_{2}$ and $H I$ are $0.5 \times 1 0^{- 3}$ and $8 \times 1 0^{- 3} M$ respectively. The equilibrium concentration of $H_{2}$ is

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

$0.5 \times 1 0^{- 3} M$

$2 \times 1 0^{- 3} M$

$4 \times 1 0^{- 3} M$

$H_{2} + I_{2} ⇌ 2 H I$

$K = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$

$[H_{2}] = \frac{[ H I ] ^{2}}{[ I _{2} ] \times K} = \frac{( 8 \times 1 0 ^{- 3} ) ^{2}}{( 0.5 \times 1 0 ^{- 3} ) ( 32 )}$

$= 4 \times 1 0^{- 3} M$

Q. The equilibrium constant for the reaction, H2​(g)+I2​(g)⇌2HI(g) is 32 at a given temperature. The equilibrium concentrations of I2​ and HI are 0.5×10−3 and 8×10−3M respectively. The equilibrium concentration of H2​ is

1×10−3M

0.5×10−3M

2×10−3M

4×10−3M