For the reaction, SO2+12O2⇌SO3, if we write Kp=Kc(RT)x, then x becomes:

Given:SO2(g)+12O2(g)⇌SO3(g) 1mol 12mol 1mol; Δng=1−1+12=−12Since, Kp=Kc(RT)x, we have x=Δng=−12 explained

For the reaction, SO2+12O2⇌SO3, if we write Kp=Kc(RT)x, then x becomes:

Given:SO2(g)+12O2(g)⇌SO3(g) 1mol 12mol 1mol; Δng=1−1+12=−12Since, Kp=Kc(RT)x, we have x=Δng=−12 explained

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For the reaction, ${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3},$ if we write $K_{p} = K_{c} (RT)^{x}$ , then x becomes:

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$- 1$

$- \frac{1}{2}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

Given:

${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g)$

$1mol \frac{1}{2} mol 1 mol; {Δn}_{g} = 1 - (1 + \frac{1}{2}) = \frac{- 1}{2}$

Since, $K_{p} = K_{c} (RT)^{x}, we have x = {Δn}_{g} = \frac{- 1}{2}$ explained

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For the reaction, SO2+12O2⇌SO3, if we write Kp=Kc(RT)x, then x becomes:

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For the reaction, ${SO}_{2} + \frac{1}{2} O_{2} ⇌ {SO}_{3},$ if we write $K_{p} = K_{c} (RT)^{x}$ , then x becomes: