The equilibrium constant for the reaction A( g )+2 B( g ) arrow C( g ) is 0.25 dm 6 mole -2. In a volume of 5 dm 3, what amount of A must be mixed with 4 moles of B to yield 1 mole of C at equilibrium.

Question

The equilibrium constant for the reaction A( g )+2 B( g ) arrow C( g ) is 0.25 dm 6 mole -2. In a volume of 5 dm 3, what amount of A must be mixed with 4 moles of B to yield 1 mole of C at equilibrium. (A) 12.5 moles (B) 26 moles (C) 25 moles (D) 13 moles

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/b735b4967be3a0518095967f9d7d6c77-.png" /> k C =0.25=([ C ]/[ A ][ B ]2) ⇒[ A ]=([ C ]/[ B 2](0.25)) (n-1/5)=(1/5 × 0.25 × (22)52)=25 ⇒ n=26 mols

Q. The equilibrium constant for the reaction $A_{(g)} + 2 B_{(g)} \to C_{(g)}$ is $0.25 d m^{6} m o l e^{- 2}$ . In a volume of $5 d m^{3}$ , what amount of A must be mixed with 4 moles of $B$ to yield 1 mole of $C$ at equilibrium.

12.5 moles

26 moles

25 moles

13 moles

$k_{C} = 0.25 = \frac{[ C ]}{[ A ] [ B ] ^{2}} \Rightarrow [A] = \frac{[ C ]}{[ B ^{2} ] ( 0.25 )}$
$\frac{n - 1}{5} = \frac{1}{5 \times 0.25 \times \frac{2 ^{2}}{5 ^{2}}} = 25 \Rightarrow n = 26$ mols

Q. The equilibrium constant for the reaction A(g)​+2B(g)​→C(g)​ is 0.25dm6mole−2. In a volume of 5dm3, what amount of A must be mixed with 4 moles of B to yield 1 mole of C at equilibrium.