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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

At 700K, eq...

Question

At $700 K$ , equilibrium constant for the reaction: $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ is $64$ . If $0.5 m o l$ $L^{- 1}$ of $H I (g)$ is present at equilibrium at $700 K$ , what are the concentration of $H_{2} (g)$ and $I_{2} (g)$ assuming that we initially started with $H I (g)$ and allowed it to reach equilibrium at $700 K$ ?

0.0625 M

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0.034 M

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0.136 M

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0.68 M

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Solution

The correct option is C $0.0625 M$
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) K = 64, T = 700 K$
$2 H I ⇌ H_{2} + I_{2} K = 1 / 64700 K$
$a$ $0$ $0$
$a (1 - α)$ $a α / 2$ $a α / 2$
$0.5$ $x$ $x$ $x$ be concentration
$\frac{x^{2}}{{(0.5)}^{2}} = \frac{1}{64} x = 1 / 16$
$x = 0.0625$

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Similar questions

Q. At

700 K

, the equilibrium constant for the reaction;

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}

54.8

0.5 m o l l i t r e^{- 1}

H I_{(g)}

is present at equilibrium at

700 K

, what are the concentrations of

H_{2 (g)}

and

I_{2 (g)}

assuming that we initially started with

H I_{(g)}

and allowed it to reach equilibrium at

700 K

At 700 K, equilibrium constant for the reaction

is 54.8. If 0.5 molL^–1 of HI₍_g₎ is present at equilibrium at 700 K, what are the concentration of H₂₍_g₎ and I₂₍_g₎ assuming that we initially started with HI₍_g₎ and allowed it to reach equilibrium at 700 K?

At 700 K, equilibrium constant for the reaction

Q. At 700 K, equilibrium constant for the reaction:

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

is 64. If

0.5 m o l L^{- 1}

of HI(g) is present at equilibrium at 700 K, what are the concentration of

H_{2} (g)

and

I_{2} (g)

assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Q. At

500

K, the equilibrium constant for the reaction

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(} g) i s 24.8

. If

\frac{1}{2} m o l / L

of HI is present at equilibrium. What are the concentrations of

H_{2}

and

I_{2}

, assuming that we started by taking HI and reached the equilibrium at

500

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At 700K, equilibrium constant for the reaction: H2(g)+I2(g)⇌2HI(g) is 64. If 0.5mol L−1 of HI(g) is present at equilibrium at 700K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

The correct option is C 0.0625MH2(g)+I2(g)⇌2HI(g)K=64,T=700K 2HI⇌H2+I2K=1/64700K a 0 0a(1−α) aα/2 aα/2 0.5 x x x be concentrationx2(0.5)2=164x=1/16x=0.0625

The correct option is C $0.0625 M$
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) K = 64, T = 700 K$
$2 H I ⇌ H_{2} + I_{2} K = 1 / 64700 K$
$a$ $0$ $0$
$a (1 - α)$ $a α / 2$ $a α / 2$
$0.5$ $x$ $x$ $x$ be concentration
$\frac{x^{2}}{{(0.5)}^{2}} = \frac{1}{64} x = 1 / 16$
$x = 0.0625$