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For the reaction, $$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_{2}(g); \:K_{c}$$ at $$427^{\circ}$$ is $$3\times 10^{-6}$$. The value of $$K_{p}$$ is nearly:


A
7.50×105
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B
2.50×105
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C
2.50×104
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D
1.72×104
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Solution

The correct option is D $$1.72 \times 10^{-4}$$
The relation between $$K_p$$ and $$K_c$$ is as follows:
$$K_{P} = K_{C} (RT)^{\Delta n }$$
$$\Delta n= $$change in gaseous moles.
For the reaction, $$2NOCl(g)\rightleftharpoons 2NO(g) + Cl_{2}(g)$$,
$$K_{C}=3 \times 10^{-6}$$ L/mol, $$R = 0.0821$$  Latm/K mol, $$\Delta n=3-2=1$$
$$T= 427 + 273 = 700\: K$$
$$K_{P}= 3 \times 10^{-6} (0.0821 \times 700)^{1}$$ $$= 1.72 \times 10^{-4}$$.
Therefore, the value of $$K_p$$ is $$1.72 \times 10^{-4}$$.

Chemistry

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