In the following gaseous phase equilibrium at constant temperature- 2 SO 2 g - O 2 g - 2 SO 3 g The concentration of - SO 2-3-0 - 10-3 M - O 2-3-5 - 10-3 M - SO 3-5-0 - 10-2 M- Calculate equilibrium constant-A- Kc-7-936 - 104B- K c -2-936 - 104C- K c -17-936 - 104D- K c -12-936 - 102

The correct option is A K_c = 7.936 × 10^4 nbsp;Given, nbsp;the concentration of [SO_2] = 3.0 × 10^ 3 M; [O_2] = 3.5 × 10^ 3M ; [SO_3] = 5.0 × 10^ 2M For ...

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Standard XII

Chemistry

Derivation of Kp and Kc

In the follow...

Question

In the following gaseous phase equilibrium at constant temperature.
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
The concentration of $[S O_{2}] = 3.0 \times 10^{- 3} M, [O_{2}] = 3.5 \times 10^{- 3} M, [S O_{3}] = 5.0 \times 10^{- 2} M$ . Calculate equilibrium constant.

K_{c} = 7.936 \times 10^{4}

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K_{c} = 17.936 \times 10^{4}

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K_{c} = 12.936 \times 10^{2}

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K_{c} = 2.936 \times 10^{4}

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Solution

The correct option is A $K_{c} = 7.936 \times 10^{4}$
Given, the concentration of $[S O_{2}] = 3.0 \times 10^{- 3} M; [O_{2}] = 3.5 \times 10^{- 3} M; [S O_{3}] = 5.0 \times 10^{- 2} M$
For the reaction,
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
Putting all the values,
$= \frac{(5 \times 10^{- 2})^{2}}{(3 \times 10^{- 3})^{2} (3.5 \times 10^{- 3})} = 7.936 \times 10^{4}$

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Similar questions

Q. In the following gaseous phase equilibrium at constant temperature.

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

The concentration of

[S O_{2}] = 3.0 \times 10^{- 3} M, [O_{2}] = 3.5 \times 10^{- 3} M, [S O_{3}] = 5.0 \times 10^{- 2} M

. Calculate equilibrium constant.

Q. The following equilibrium concentrations were measured at

800 K : [S O_{2}] = 3.0 \times 10^{- 3} M; [O_{2}] = 3.5 \times 10^{- 3} M; [S O_{3}] = 5.0 \times 10^{3} M

Calculate the equilibrium constant at

800 K

for the chemical reaction in equilibrium for both directions.

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

(

K_{c}

is the equilibrium constant for the forward reaction and

K_{c}^{^{'}}

is the equilibrium constant for the backward reaction)

Q. When the molar concentrations of

S O_{2}, O_{2}

and

S O_{3}

at equilibrium at certain temperature are

0.5, 0.25, a n d 0.25 M

respectively,

K_{c}

for

S O_{3 (g)} ⇌ 2 S O_{2 (g)} + O_{2 (g)}

is:

Q. For the reaction,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

What is

K_{c}

when the equilibrium concentration of

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M a n d [S O_{3}] = 1.90 M

What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?

2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

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Derivation of Kp and Kc

Standard XII Chemistry

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In the following gaseous phase equilibrium at constant temperature. 2SO2(g)+O2(g)⇌2SO3(g) The concentration of [SO2]=3.0×10−3M, [O2]=3.5×10−3 M, [SO3]=5.0×10−2 M. Calculate equilibrium constant.

The correct option is A Kc=7.936×104 Given, the concentration of [SO2]=3.0×10−3M; [O2]=3.5×10−3M; [SO3]=5.0×10−2M For the reaction, 2SO2(g)+O2(g)⇌2SO3(g) Kc=[SO3]2[SO2]2[O2] Putting all the values, =(5×10−2)2(3×10−3)2(3.5×10−3)=7.936×104

In the following gaseous phase equilibrium at constant temperature.
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
The concentration of $[S O_{2}] = 3.0 \times 10^{- 3} M, [O_{2}] = 3.5 \times 10^{- 3} M, [S O_{3}] = 5.0 \times 10^{- 2} M$ . Calculate equilibrium constant.