Chemistry, asked by LoverLoser, 26 days ago

SOLVE THE CHEMISTRY QUESTION
*EQUILIBRIUM*

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Answers

Answered by snehitha2
11

Question:

For the equilibrium  \sf AB_{(g)} \rightleftharpoons A_{(g)}+B_{(g)} , \rm K_p is equal to four times the total pressure. The number of moles of 'A' formed is

Answer:

option (4) 2/√5

Explanation:

Let the total equilibrium pressure be P atm

\rm K_p is four times the total pressure.

\longrightarrow \sf K_p=4P

Let the degree of dissociation be α

\sf AB_{(g)} \rightleftharpoons A_{(g)}+B_{(g)}

  1           0        0      ( initially)

1 - α         α        α      (at equilibrium)

Total number of moles at equilibrium = 1 - α + α + α

      = 1 + α

  • The partial pressure is the pressure exerted by an individual gas in a mixture of gases.

Now, let's find the partial pressures of AB, A and B

Partial pressure of AB :

\sf P_{AB}=\dfrac{1-\alpha}{1+\alpha} \times P

Partial pressure of A :

\sf P_{A}=\dfrac{\alpha}{1+\alpha} \times P

Partial pressure of B :

\sf P_{B}=\dfrac{\alpha}{1+\alpha} \times P

The expression of  \rm K_p for the given equation is

\boxed{\bf K_p=\dfrac{P_A \times P_B}{P_{AB}}}

Substitute,

 \sf 4P=\dfrac{\bigg(\dfrac{\alpha}{1+\alpha} \times P \bigg) \times \bigg(\dfrac{\alpha}{1+\alpha} \times P \bigg)}{\dfrac{1-\alpha}{1+\alpha} \times P} \\\\ \sf 4P=\dfrac{\dfrac{\alpha^2}{(1+\alpha)^2} \times P^2}{\dfrac{1-\alpha}{1+\alpha} \times P} \\\\ \sf 4=\dfrac{\dfrac{\alpha^2}{(1+\alpha)}}{1-\alpha} \\\\ \sf 4=\dfrac{\alpha^2}{(1-\alpha)(1+\alpha)} \\\\ \sf 4=\dfrac{\alpha^2}{1^2-\alpha^2} \\\\ \sf 4(1-\alpha^2)=\alpha^2 \\\\ \sf 4-4\alpha^2=\alpha^2 \\\\ \sf 4\alpha^2+\alpha^2=4

 \sf 5\alpha^2=4 \\\\ \sf \alpha^2=\dfrac{4}{5} \\\\ \tt \alpha=\sqrt{\dfrac{4}{5}} \\\\ \longrightarrow \tt \alpha=\dfrac{2}{\sqrt{5}}

The number of moles of 'A' formed is 2/√5

Answered by aryansaini25
0

Answer:

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Explanation:

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