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Standard XII

Chemistry

Derivation of Kp and Kc

The value of ...

Question

The value of $K_{p}$ for the reaction
$2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)$ is 0.03 atm at $427^{\circ} C$ , when the partial pressure are expressed in atmosphere, then the value of $K_{c}$ for the same reaction is

5.23 \times 10^{- 4}

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7.34 \times 10^{- 4}

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3.2 \times 10^{- 3}

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5.43 \times 10^{- 5}

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Solution

The correct option is A $5.23 \times 10^{- 4}$
For the given reaction,
$Δ n_{g} = 4 + 1 - (2 + 2) = 1 W e k n o w, K_{p} = K_{c} (R T)^{Δ n_{a}} G i v e n, K_{p} = 0.03 a n d T = 700 K ∴ 0.03 = K_{c} (0.082 \times 700)^{1} K_{c} = 5.23 \times 10^{- 4}$

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Similar questions

Q. The value of

K_{p}

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2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)

is 0.03 atm at

427^{\circ} C

, when the partial pressure are expressed in atmosphere, then the value of

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Q. The value of

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Q. For the reversible reaction,

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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Derivation of Kp and Kc

Standard XII Chemistry

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The value of Kp for the reaction 2H2O(g)+2Cl2(g)⇌4HCl(g)+O2(g) is 0.03 atm at 427∘C, when the partial pressure are expressed in atmosphere, then the value of Kc for the same reaction is

The correct option is A 5.23×10−4For the given reaction, Δng=4+1−(2+2)=1We know,Kp=Kc(RT)ΔnaGiven,Kp=0.03 and T=700 K∴0.03=Kc(0.082×700)1Kc=5.23×10−4

The value of $K_{p}$ for the reaction
$2 H_{2} O (g) + 2 C l_{2} (g) ⇌ 4 H C l (g) + O_{2} (g)$ is 0.03 atm at $427^{\circ} C$ , when the partial pressure are expressed in atmosphere, then the value of $K_{c}$ for the same reaction is

The correct option is A $5.23 \times 10^{- 4}$
For the given reaction,
$Δ n_{g} = 4 + 1 - (2 + 2) = 1 W e k n o w, K_{p} = K_{c} (R T)^{Δ n_{a}} G i v e n, K_{p} = 0.03 a n d T = 700 K ∴ 0.03 = K_{c} (0.082 \times 700)^{1} K_{c} = 5.23 \times 10^{- 4}$