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Derivation of Kp and Kc

For the react...

Question

For the reaction ${S O}_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ {S O}_{3 (g)}$ , if $K_{P} = K_{C} {(R T)}^{x}$ where the symbols have usual meaning then the value of $x$ is: (assuming ideality)

- 1

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- \frac{1}{2}

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\frac{1}{2}

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1

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Solution

The correct option is C $- \frac{1}{2}$
For the reaction:

${S O}_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ {S O}_{3 (g)}$ ,
$K_{P} = K_{C} {(R T)}^{x}$

$x$ =moles of gases in product-moles of gases in reactant

$x = 1 - (1 - 1 / 2)$

$x = - 1 / 2$

Option B is correct.

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Similar questions

S O_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ S O_{3 (g)}

K_{P} = K_{C} (R T)^{x}

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S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)

K_{c} = K_{p} (R T)^{x}

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S O_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ S O_{3 (g)}

K_{P} = K_{C} (R T)^{x}

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S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ S O_{3} (g)

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K_{1}

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S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g) 2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

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