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Stoichiometry

A vessel of 2...

Question

A vessel of 250 litre was filled with 0.01 mole of $S b_{2} S_{3}$ and 0.01 mole of $H_{2}$ to attain the equilibrium at $440^{\circ}$ C as $S b_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)$ .
After equilibrium the $H_{2}$ S formed was analysed by dissolving it in water and treating with excess of $P b^{2 +}$ to give 1.195 g of PbS (Molecular weight=239)precipitate.
What is value of $K_{c}$ of the reaction at $440^{\circ}$ C?

A

1

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B

2

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C

4

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D

None of these

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Solution

The correct option is A 1
1.195 g of PbS (Molecular weight=239) precipitate is formed. Hence, $[H_{2} S] = \frac{1.195}{239} = 0.005 m o l e s$ . The equilibrium concentration of hydrogen sulfide is $\frac{0.005}{250} M$ .
The equilibrium number of moles of hydrogen is $0.01 - 0.005 = 0.005 m o l e s$ .
The equilibrium concentration of hydrogen is $[H_{2}] = \frac{0.005}{250} M$ .
The expression for the equilibrium constant is $K_{c} = (\frac{[H_{2} S]}{H_{2}})^{3} = (\frac{\frac{0.005}{250}}{\frac{0.005}{250}})^{3} = 1$ .

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Similar questions

Q. A vessel of 250 litre was filled with 0.01 mole of

S b_{2} S_{3}

and 0.01 mole of

H_{2}

to attain the equilibrium at

440 ° C

S b_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)

.
After equilibrium, the

H_{2} S

formed was analysed by dissolving it in water and treating with excess of

P b^{2 +}

1.195 g

P b S

as precipitate. What is the value of

K_{C}

440 ° C

? (Molar mass of

P b = 207 g m o l^{- 1}

250

litre was filled with

0.01

{S b}_{2} S_{3}

0.01

H_{2}

to attain the equilibrium at

440^{o} C

{S b}_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)

After equilibrium, the

H_{2} S

formed was analysed by dissolved it in water and treating with excess of

{P b}^{2 +}

1.19

P b S

as precipitate. What is the value of

K_{c}

440^{o} C

.

Q. A vessel of 250 litre was filled with 0.01 mole of

S b_{2} S_{3}

and 0.01 mole of

H_{2}

to attain the equilibrium at

440^{\circ} C

as given below.

S b_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)

.
In a separate experiment, this

H_{2} S

is sufficient to precipitate 1.19 g

P b S

P b^{2 +}

solution. Calculate

K_{c}

of the above reaction.

2.50 l i t r e

was filled with

0.01

S b_{2} S_{3}

0.01

H_{2}

to attain the equilibrium at

440

S b_{2} S_{3} (s) + 3 H_{2} (s) ⇌ 2 S b (s) + 3 H_{2} S (g)

.
After equilibrium the

H_{2} S

formed was analysed by dissolving it in water and treating with excess of

P b^{2 +}

1.029 g

P b S

as precipitate. The value of

K_{c}

of the reaction at

440

X

P b = 206

100 X

A mixture of $H_{2}$ and $I_{2}$ in molecular proportion of 2 : 3 was heated at $440^{\circ} C$ till the reaction $H_{2} + I_{2}$ $⇌$ 2HI reached equilibrium state. Calculate the percentage of iodine converted into HI $K_{c} a t 440$ $^{\circ}$ C is 0.02)

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