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Reaction Quotient

The value of ...

Question

The value of $K_{p}$ for dissociation of

$2 H I (g) ⟺ H_{2} (g) + I_{2} (g)$

is $1.84 \times 10^{- 2}$ . If the equilibrium concentration of $H_{2}$ is 0.4789 mol litre $^{- 1}$ , calculate the concentration of $H I$ at equilibrium.

3.53 mol lit

^{- 1}

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0.53 mol lit

^{- 1}

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35.3 mol lit

^{- 1}

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0.353 mol lit

^{- 1}

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Solution

The correct option is A 3.53 mol lit $^{- 1}$
As $Δ n = 0$ , $K_{p} = K_{c} = 1.84 \times 10^{- 2}$ .

Also at equilibrium, $[H_{2}] [I_{2}] = 0.4789 M$ .

The expression for the equilibrium constant becomes $K_{c} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}}$ .

Substitute values in the above expression.

$1.84 \times 10^{- 2} = \frac{0.4789 \times 0.4789}{[H I]^{2}}$ .

$[H I] = 3.53 m o l l i t^{- 1}$ .

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The value of Kp for dissociation of 2HI(g)⟺H2(g)+I2(g) is 1.84×10−2. If the equilibrium concentration of H2 is 0.4789 mol litre−1, calculate the concentration of HI at equilibrium.

The correct option is A 3.53 mol lit−1As Δn=0, Kp=Kc=1.84×10−2. Also at equilibrium, [H2][I2]=0.4789M.The expression for the equilibrium constant becomes Kc=[H2][I2][HI]2.Substitute values in the above expression.1.84×10−2=0.4789×0.4789[HI]2.[HI]=3.53 mol lit−1.

The value of $K_{p}$ for dissociation of

$2 H I (g) ⟺ H_{2} (g) + I_{2} (g)$

is $1.84 \times 10^{- 2}$ . If the equilibrium concentration of $H_{2}$ is 0.4789 mol litre $^{- 1}$ , calculate the concentration of $H I$ at equilibrium.