For the reaction in equilibrium I COg - 1-2 O2 g - CO2 g II 2COg - O2 g - 2CO2 g Then-Equilibrium constant Free energy changeK1-G1K2-G2

For the reaction, CO(g) + 1/2 nbsp; O_2 (g) ⇌ CO_2 (g) nbsp; ...(1) K_1=[CO_2]/[CO][O_2]^1/2 Again for the reaction, 2CO(g) + O_2 (g) ⇌ 2CO_2 (g) K_2=[CO_2] ...

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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

For the react...

Question

For the reaction in equilibrium
(I) $C O (g) + \frac{1}{2} O_{2} (g) ⇌ C O_{2} (g)$
(II) $2 C O (g) + O_{2} (g) ⇌ 2 C O_{2} (g)$
Then,

Equilibrium constant Free energy change

$K_{1}$ $Δ G_{1}$

$K_{2}$ $Δ G_{2}$

K_{2} = K_{1}^{2}, Δ G_{2} = 2 Δ G_{1}

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K_{2} = 2 K_{1}, Δ G_{2} = 2 Δ G_{1}

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K_{1} = 2 K_{2}, Δ G_{1} = 2 Δ G_{1}

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K_{2} = K_{1}^{2}, Δ G_{2} = 2 Δ G_{1}^{2}

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Solution

The correct option is A $K_{2} = K_{1}^{2}, Δ G_{2} = 2 Δ G_{1}$
For the reaction,
$C O (g) + \frac{1}{2} O_{2} (g) ⇌ C O_{2} (g)$ ...(1)
$K_{1} = \frac{[C O_{2}]}{[C O] [O_{2}]^{\frac{1}{2}}}$
Again for the reaction,
$2 C O (g) + O_{2} (g) ⇌ 2 C O_{2} (g)$
$K_{2} = \frac{[C O_{2}]^{2}}{[C O]^{2} [O_{2}]} = K_{1}^{2}$

We know,
$Δ G_{1} = - 2.303 R T l o g K_{1}$
$Δ G_{2} = - 2.303 R T l o g K_{2} = 2.303 R T l o g K_{1}^{2}$
$\Rightarrow Δ G_{2} = 2 \times (- 2.303 R T l o g K_{1}) = 2 Δ G_{1}$

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Similar questions

Q. For the reaction in equilibrium
(I)

C O (g) + \frac{1}{2} O_{2} (g) ⇌ C O_{2} (g)

(II)

2 C O (g) + O_{2} (g) ⇌ 2 C O_{2} (g)

Then,

Equilibrium constant	Free energy change
$K_{1}$	$Δ G_{1}$
$K_{2}$	$Δ G_{2}$

Q. Calculate the equilibrium constant for the reaction

H_{2 (g)} + C O_{2 (g)} ⇌ H_{2} O_{(g)} + C O_{(g)}

1395 K

, if the equilibrium constants at

1395 K

for following are:

2 H_{2} O_{(g)} ⇌ 2 H_{2} + O_{2 (g)}; K_{1} = 2.1 \times 10^{- 13}

2 C O_{2 (g)} ⇌ 2 C O_{(g)} + O_{2 (g)}; K_{2} = 1.4 \times 10^{- 12}

Q. The equilibrium constants for the following reactions at

1400 K

are given:

2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g); K_{1} = 2.1 \times 10^{- 13}

2 {C O}_{2} (g) ⇌ 2 C O (g) + O_{2} (g); K_{2} = 1.4 \times 10^{- 12}

Then, the equilibrium constant

K

for the reaction,

H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g)

, is:

Q. At 1000 K, the equilibrium constants for the following equilibrium are

K_{1}

and

K_{2}

C O C l_{2} (g) ⇌ C O (g) + C l_{2} (g)

;

K_{1} = 0.329

2 C O (g) + O_{2} (g) ⇌ 2 C O_{2} (g)

;

K_{2} = 2.24 \times 10^{22}

For the equilibrium,

2 C O C l_{2} (g) + O_{2} (g) ⇌ 2 C O_{2} (g) + 2 C l_{2} (g)

, the equilibrium constant

K_{3}

is:

Q. Equilibrium constants for the following reaction at 1200 K are given.

2 H_{2} O (g) ⇋ 2 H_{2} (g) + O_{2} (g); K_{1} = 6.4 \times 10^{- 8}

2 C O_{2} (g) ⇋ 2 C O (g) + O_{2} (g); K_{2} = 1.6 \times 10^{- 6}

The equilibrium constant for the reaction

H_{2} (g) + C O_{2} (g) ⇋ C O (g) + H_{2} O (g)

at 1200 K will be:

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For the reaction in equilibrium (I) CO(g)+12O2(g)⇌CO2(g) (II) 2CO(g)+O2(g)⇌2CO2(g) Then, Equilibrium constant Free energy change K1 ΔG1 K2 ΔG2

The correct option is A K2=K21, ΔG2=2ΔG1For the reaction, CO(g)+12O2(g)⇌CO2(g) ...(1) K1=[CO2][CO][O2]12 Again for the reaction, 2CO(g)+O2(g)⇌2CO2(g) K2=[CO2]2[CO]2[O2]=K21 We know, ΔG1=−2.303RTlogK1 ΔG2=−2.303RTlogK2=2.303RTlogK21 ⇒ΔG2=2×(−2.303RTlogK1)=2ΔG1

For the reaction in equilibrium
(I) $C O (g) + \frac{1}{2} O_{2} (g) ⇌ C O_{2} (g)$
(II) $2 C O (g) + O_{2} (g) ⇌ 2 C O_{2} (g)$
Then,

Equilibrium constant Free energy change

$K_{1}$ $Δ G_{1}$

$K_{2}$ $Δ G_{2}$