Consider the following gaseous equilibrium in a closed container of volume “V” at T(K). P2(g) + Q2(g) 2PQ(g) 2 moles each of P2(g), Q2(g) and PQ (g) are present at equilibrium. Now one mole each of ‘P

Question

Consider the following gaseous equilibrium in a closed container of volume “V” at T(K). P2(g) + Q2(g) 2PQ(g) 2 moles each of P2(g), Q2(g) and PQ (g) are present at equilibrium. Now one mole each of ‘P2’ and ‘Q2’ are added to the equilibrium keeping the temperature at T(K). The number of moles of P2, Q2 and PQ at the new equilibrium, respectively , are -

(A) 2.67, 2.67, 2.67
(B) 1.21, 2.24, 1.56
(C) 1.66, 1.66, 1.66
(D) 2.56, 1.62, 2.24

Prashant Kotian · Accepted Answer

Correct answer: A. P2(g) + Q2(g) 2PQ(g) t=teq 2 mole 2 mole 2 mole Keq = 2 2.2  Now 1 mole of each P2 and Q2 is added So reaction will move in forward direction P2(g) + Q2(g) 2PQ(g) eq. t t'  3 – x 3 – x 2 + 2x KC = 1 = (2 2x) (3 x)(3 x)    2x 3 x    2 + 2x = 3 – x x = 1 At new equilibrium : Moles of P2 = 8 3 = 2.67 Moles of Q2 = 8 3 = 2.67 Moles of PQ = 8 3 = 2.67

💡 Solution & Explanation

Practice 22,000+ questions like this