2 HX g - H 2 g - X 2 g - K C -1-0 - 10-5 what is the concentration of HX if the equilibrium concentration of H 2 and X 2 is 1-2 -10-3 M and 1-2 - 10-4 M respectively-A- 1-2 - 10-4 M- 1-2 - 10-3 MC- 1-2 - 10-2 MD- 1-2 - 10-1 M

The correct option is D 1.2 × 10^ 1 MGiven, 2HX ⇋ H_2 (g) + X_2 (g) ; K_C = 1.0 × 10^ 5 ⇒ K_C = [H_2][X_2]/[HX]^2 [HX]^2 = [H_2][X_2]/K_C [HX]^2 = 1.2 × 10^ ...

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Byju's Answer

Standard XII

Chemistry

pH the Power of H

2 HX g ⇋ H 2 ...

Question

$2 H X (g) ⇋ H_{2} (g) + X_{2} (g); K_{C} = 1.0 \times 10^{- 5}$ what is the concentration of $H X$ if the equilibrium concentration of $H_{2}$ and $X_{2}$ is $1.2 \times 10^{- 3} M$ and $1.2 \times 10^{- 4} M$ respectively?

1.2 \times 10^{- 4} M

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1.2 \times 10^{- 3} M

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1.2 \times 10^{- 2} M

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1.2 \times 10^{- 1} M

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Solution

The correct option is D $1.2 \times 10^{- 1} M$
Given,
$2 H X ⇋ H_{2} (g) + X_{2} (g); K_{C} = 1.0 \times 10^{- 5}$
$\Rightarrow K_{C} = \frac{[H_{2}] [X_{2}]}{[H X]^{2}}$
$[H X]^{2} = \frac{[H_{2}] [X_{2}]}{K_{C}}$
$[H X]^{2} = \frac{1.2 \times 10^{- 3} \times 1.2 \times 10^{- 4}}{1.0 \times 10^{- 5}}$
$∴ H X = 1.2 \times 10^{- 1} M$

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Similar questions

2 H X (g) ⇋ H_{2} (g) + X_{2} (g); K_{C} = 1.0 \times 10^{- 5}

What is the concentration of

H X

if the equilibrium concentration of

H_{2}

and

X_{2}

1.2 \times 10^{- 3} M

and

1.2 \times 10^{- 4} M

respectively?

Q. For the equilibrium system

2 H X (g) ⇌ H_{2} (g) + X_{2} (g)

the equilibrium constant is

1.0 \times 10^{- 5}

. What is the concentration of HX if the equilibrium concentration of

H_{2}

and

X_{2}

are

1.2 \times 10^{- 3} M

, and

1.2 \times 10^{- 4} M

respectively?

Q. Calculate the equilibrium constant

(K c)

for the formation of

N H_{3}

in the following reaction:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

At equilibrium, the concentration of

N H_{3}, H_{2}

and

N_{2}

are

1.2 \times 10^{- 2}, 3.0 \times 10^{- 2}

and

1.5 \times 10^{- 2} M

respectively.

Q. The

K_{s p}

for

M g (O H)_{2}

in water is

1.2 \times 10^{- 11}

. In an acidic solution, if the

M g^{2 +}

concentration is

1.2 \times 10^{- 5} m o l / L

, what is the pH at which

M g (O H)_{2}

just begins to precipitate?

Q. The following concentration were obtained for the formation of

N H_{3}

form

N_{2}

and

H_{2}

at equilibrium for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

[N_{2}] = 1.5 \times 10^{- 2} M

[H_{2}] = 3.0 \times 10^{- 2} M

[N H_{3}] = 1.2 \times 10^{- 2} M

Calculate equilibrium constant.

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2HX(g)⇋H2(g)+X2(g); KC=1.0×10−5 what is the concentration of HX if the equilibrium concentration of H2 and X2 is 1.2×10−3 M and 1.2×10−4 M respectively?

The correct option is D 1.2×10−1 MGiven, 2HX⇋H2(g)+X2(g); KC=1.0×10−5 ⇒KC=[H2][X2][HX]2 [HX]2=[H2][X2]KC [HX]2=1.2×10−3×1.2×10−41.0×10−5 ∴HX=1.2×10−1 M

$2 H X (g) ⇋ H_{2} (g) + X_{2} (g); K_{C} = 1.0 \times 10^{- 5}$ what is the concentration of $H X$ if the equilibrium concentration of $H_{2}$ and $X_{2}$ is $1.2 \times 10^{- 3} M$ and $1.2 \times 10^{- 4} M$ respectively?