$1.50$ mol each of hydrogen and iodine were placed in a sealed $10 L$ container maintained at $717 K$ . At equilibrium $1.25$ mol each of hydrogen and iodine were left behind. The equilibrium constant, $K_{c}$ for the reaction.
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ at $717 K$ is

0.4

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0.16

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25

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50

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Solution

The correct option is C $0.16$
$H_{2} + I_{2} ⇋ 2 H I$
$1.5$ $1.5$ $0$
$1.25$ $1.25$ $0.5$
$1.5 - x$ $1.5 - x$ $2 x$
$1.5 - x = 1.25$
$⟹ x = 0.25$
So, moles of $H I$ produced $= 0.5$ $m o l e s$
So, $[H I] = \frac{0.5}{10} = 0.05$ $m o l / L$ ; $[H_{2}] = \frac{1.25}{10} = 0.125$
$[I_{2}] = \frac{1.25}{10} = 0.125$
$⟹ K_{C} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]} = \frac{(0.05)^{2}}{(0.125) (0.125)} = \frac{{(\frac{1}{20})}^{2}}{\frac{1}{8} \cdot \frac{1}{8}} = \frac{64}{(20)^{2}} = 0.16$

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Similar questions

1.5 Moles each of hydrogen and Iodine were placed in a sealed 10L container maintained at 717K . At equilibrium 1.25 moles each of hydrogen and Iodine were left behind .The equilibrium constant K_cfor the reaction H₂+I_{2↔ 2HI} gives at 717K.

Q. 1.50 mol each of hydrogen and iodine were placed in a sealed 10 L container maintained at 717 K. At equilibrium 1.25 mol each of hydrogen and iodine were left behind. The equilibrium constant,

K_{c}

for the reaction

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) a t 717 K

4.5

moles each of hydrogen and iodine heated in a sealed ten-litre vessel. At equilibrium,

3

moles of

H I

were found.

The equilibrium constant for

H_{2} (g)

I_{2} (g) ⇌ 2 H I (g)

is:

Q. 4.5 moles each of hydrogen and iodine are heated in a sealed 10-liter vessel. At equilibrium 3.0 moles of HI were found.

The equilibrium constant for

H_{2} (g)

I_{2} (g) ⇌ 2 H I (g)

is:

Q. 4.5 moles of each hydrogen and iodine are heated in a sealed 10-liter vessel. At equilibrium, 3 moles of hydrogen iodide was found.

The equilibrium constant for

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is;

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1.50 mol each of hydrogen and iodine were placed in a sealed 10 L container maintained at 717 K. At equilibrium 1.25 mol each of hydrogen and iodine were left behind. The equilibrium constant, Kc for the reaction.H2(g)+I2(g)⇌2HI(g) at 717 K is

The correct option is C 0.16 H2+I2⇋2HI 1.5 1.5 0 1.25 1.25 0.51.5−x 1.5−x 2x1.5−x=1.25⟹x=0.25So, moles of HI produced =0.5 molesSo, [HI]=0.510=0.05 mol/L; [H2]=1.2510=0.125[I2]=1.2510=0.125⟹KC=[HI]2[H2][I2]=(0.05)2(0.125)(0.125)=(120)218⋅18=64(20)2=0.16

$1.50$ mol each of hydrogen and iodine were placed in a sealed $10 L$ container maintained at $717 K$ . At equilibrium $1.25$ mol each of hydrogen and iodine were left behind. The equilibrium constant, $K_{c}$ for the reaction.
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ at $717 K$ is