For the equilibrium AB(g) = A(g) + B(g). Kp is
equal to four times the total pressure. The number
of moles of 'A' formed is(1)√5/2
(2) 2
(3) √5
(4)2/√5
Answers
Answer:
Explanation:
=> Reaction :
AB(g) ⇔ A(g) + B(g)
(equillibrium) (a−x) x x
=> Here, x is the number of moles at equillibrium.
=> Suppose, equillibrium pressure = P
=> Thus, The total number of moles at equillibrium is equal to a + x
∴ P_A = P_B =x/a+x * P
P_AB =a−x/a+x
K_p =P_A*P_B/P_AB
4P = (x/a+x * P) * (x/a+x *P ) * (a−x/a+x)
4a²−4x² =x²
4a² = x² + 4x²
4a² = 5x²
x² = 4a² / 5
x = 2 a / √5
Thus, the number of moles of A is 2/√5 * a.
Answer:Reaction :
AB(g) ⇔ A(g) + B(g)
(equillibrium) (a−x) x x
=> Here, x is the number of moles at equillibrium.
=> Suppose, equillibrium pressure = P
=> Thus, The total number of moles at equillibrium is equal to a + x
∴ P_A = P_B =x/a+x * P
P_AB =a−x/a+x
K_p =P_A*P_B/P_AB
4P = (x/a+x * P) * (x/a+x *P ) * (a−x/a+x)
4a²−4x² =x²
4a² = x² + 4x²
4a² = 5x²
x² = 4a² / 5
x = 2 a / √5
Thus, the number of moles of A is 2/√5 * a.
Explanation: