Kp for the reaction A leftharpoons B is 4. If initially only A is present then what will be the partial pressure of B after equilibrium ?

Question

Kp for the reaction A leftharpoons B is 4. If initially only A is present then what will be the partial pressure of B after equilibrium ? (A) 1.2 (B) 0.8 (C) 0.6 (D) 1

Tardigrade Admin · Accepted Answer

Assuming that reaction is started with 1 mol of A and at equilibrium x moles of A have reacted, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/93cbeae59d0f21cacb63c22bd11d421c-.png" /> At equibrium 1-x mol x mol Kp=(PB/PA)=4 ⇒ (x/1-x)=4 ⇒ x =0.8

Q. $K_{p}$ for the reaction $A ⇌ B$ is $4$ . If initially only A is present then what will be the partial pressure of B after equilibrium ?

1.2

0.8

0.6

1

Assuming that reaction is started with $1$ mol of A and at equilibrium $x$ moles of A have reacted,

At equibrium $1 - x m o l$ $x$ mol
$K_{p} = \frac{P _{B}}{P _{A}} = 4 \Rightarrow \frac{x}{1 - x} = 4 \Rightarrow x = 0.8$

Q. Kp​ for the reaction A⇌B is 4. If initially only A is present then what will be the partial pressure of B after equilibrium ?

1.2

0.8

0.6

1

Assuming that reaction is started with 1 mol of A and at equilibrium x moles of A have reacted, At equibrium 1−xmol x mol Kp​=PA​PB​​=4⇒1−xx​=4⇒x=0.8

Q. $K_{p}$ for the reaction $A ⇌ B$ is $4$ . If initially only A is present then what will be the partial pressure of B after equilibrium ?

Assuming that reaction is started with $1$ mol of A and at equilibrium $x$ moles of A have reacted,

At equibrium $1 - x m o l$ $x$ mol
$K_{p} = \frac{P _{B}}{P _{A}} = 4 \Rightarrow \frac{x}{1 - x} = 4 \Rightarrow x = 0.8$