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In a 0.25-lit...

Question

In a 0.25-liter tube dissociation of 4 moles of NO is took place. If its degree of dissociation is 10 %. The value of $K_{p}$ for reaction $2 N O ⇌ N_{2} + O_{2}$ is:

\frac{1}{(18)^{2}}

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\frac{1}{(8)^{2}}

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\frac{1}{(16)}

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\frac{1}{(32)}

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Solution

The correct option is A $\frac{1}{(18)^{2}}$
$K_{p}$ for reaction $2 N O ⇌ N_{2} + O_{2}$

Initial $4$ $0$ $0$
FInal $4 \times (1 - 0.1)$ $0.5 \times 0.1 \times 4$ $0.5 \times 4$
$K_{p} = \frac{(0.5 \times 0.1 \times 4) (0.5 \times 0.1 \times 4)}{(4 \times 0.9)^{2}} = (\frac{1}{18})^{2}$

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Similar questions

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Initial	$4$	$0$	$0$
FInal	$4 \times (1 - 0.1)$	$0.5 \times 0.1 \times 4$	$0.5 \times 4$

In a 0.25-liter tube dissociation of 4 moles of NO is took place. If its degree of dissociation is 10 %. The value of Kp for reaction 2NO⇌N2+O2 is:

The correct option is A 1(18)2Kp for reaction 2NO⇌N2+O2Initial 400FInal 4×(1−0.1) 0.5×0.1×4 0.5×4Kp=(0.5×0.1×4)(0.5×0.1×4)(4×0.9)2=(118)2

In a 0.25-liter tube dissociation of 4 moles of NO is took place. If its degree of dissociation is 10 %. The value of $K_{p}$ for reaction $2 N O ⇌ N_{2} + O_{2}$ is:

The correct option is A $\frac{1}{(18)^{2}}$
$K_{p}$ for reaction $2 N O ⇌ N_{2} + O_{2}$

Initial $4$ $0$ $0$
FInal $4 \times (1 - 0.1)$ $0.5 \times 0.1 \times 4$ $0.5 \times 4$
$K_{p} = \frac{(0.5 \times 0.1 \times 4) (0.5 \times 0.1 \times 4)}{(4 \times 0.9)^{2}} = (\frac{1}{18})^{2}$