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Question: For the reaction equilibrium \(2NOBr(g)\rightleftharpoons 2NO(g)+B{{r}_{2}}(g)\), if \({{P}_{B{{r}_{...

For the reaction equilibrium 2NOBr(g)⇌2NO(g)+Br2(g)2NOBr(g)\rightleftharpoons 2NO(g)+B{{r}_{2}}(g), if PBr2=P9{{P}_{B{{r}_{2}}}}=\dfrac{P}{9} at equilibrium and P is the initial total pressure, then the ratio KpP\dfrac{{{K}_{p}}}{P} is equal to:
(a) 19\dfrac{1}{9}
(b) 4441\dfrac{4}{441}
(c) 127\dfrac{1}{27}
(d) 13\dfrac{1}{3}

Explanation

Solution

Before the equilibrium reaction the pressure of NOBrNOBr will be P and pressure of NONO and Br2B{{r}_{2}} will be zero. After equilibrium the pressure of NOBrNOBrwill be P−2P′P-2P', pressure of NONOwill be 2P′2P' and pressure of Br2B{{r}_{2}}will be P′P'. The equilibrium constant at pressure can be calculated by the ratio of the product of partial pressure of products to the partial pressure of the reactant.

Complete step by step answer:
We are given a reaction in which two moles of nitrosyl bromide in equilibrium forms two moles of nitrogen oxide and one mole of bromine gas. The reaction is:
2NOBr(g)⇌2NO(g)+Br2(g)2NOBr(g)\rightleftharpoons 2NO(g)+B{{r}_{2}}(g)
So, before the equilibrium reaction, the pressure of NOBrNOBr will be P and the pressure of NONO, and Br2B{{r}_{2}} will be zero. After equilibrium the pressure of NOBrNOBrwill be P−2P′P-2P', pressure of NONOwill be 2P′2P' and pressure of Br2B{{r}_{2}}will be P′P'.
And the partial pressure of bromine PBr2{{P}_{B{{r}_{2}}}} is given as: PBr2=P9{{P}_{B{{r}_{2}}}}=\dfrac{P}{9}and this is equal to P′P'.
So, by equating this value we can get the partial pressure of NONO and NOBrNOBr.
The partial pressure of NOBrNOBr= P−2P′P-2P'
NOBr=P−2P9NOBr=P-2\dfrac{P}{9}
NOBr=7P9NOBr=\dfrac{7P}{9}
The partial pressure of NONO = 2P′2P'
NO=2 x P9NO=2\text{ x }\dfrac{P}{9}
NO=2P9NO=\dfrac{2P}{9}
So, the equilibrium constant at pressure can be calculated by the ratio of the product of partial pressure of products to the partial pressure of the reactant.
Kp=(NO)2(Br2)(NOBr)2{{K}_{p}}=\dfrac{{{(NO)}^{2}}(B{{r}_{2}})}{{{(NOBr)}^{2}}}
So, putting the value of partial pressure in the formula, we get
Kp=(2P9)2(P9)(7P9)2{{K}_{p}}=\dfrac{{{\left( \dfrac{2P}{9} \right)}^{2}}\left( \dfrac{P}{9} \right)}{{{\left( \dfrac{7P}{9} \right)}^{2}}}
Kp=4P441{{K}_{p}}=\dfrac{4P}{441}
We have to find the ratio KpP\dfrac{{{K}_{p}}}{P}:
KpP=4P441P=4441\dfrac{{{K}_{p}}}{P}=\dfrac{4P}{441P}=\dfrac{4}{441}
Therefore the correct answer is an option (b) 4441\dfrac{4}{441}.

Note: When you are writing the formula of the chemical equilibrium constant, the number of moles of the compound must be the power of the compound in the formula, otherwise the value of constant will be wrong.