Question

Question: For the reaction 2 A(9) + 3 B(9) - 2 C(9) + 2 D(9), the equilibrium constant at 298 K is 1.456x10-11. If the initial partial pressures of A, and B are 0.362 bar, and 0.653 bar, respectively, determine the partial pressure of D (in bar) once equilibrium has been reached at 298 K. (2 marks)

Answer 1

Given reaction, 2 Alg) + 3 Bl) - 20092 DIG KE 1.456x10-!! at 248k Initial: 0.362 bar 0.653 bat 0 0 Change: -24 8643 -3 +264 +26 Equilibrium: (0.362-2) 10.653-3x) 3 201 1. Given initial partial pressure of ADB are R 0.362 bar and 0.653 bar caespectively. Let at equilibrium some moles of a beaved with B to form c & Dadue to which decrease is brexure of Abe 20. Then, increase for pressure of D = 20 Increase in pressure al c = 2x delyease in paestute of B= 32 By law of mass action, as equilibrium 393 putting in above egr values of k and PA PB Pe & Po we gee (26) 202x)? 1.456 *10-11 (0.362-2x)?*(0.653-3) • Left hand side of ear is very small we can igonke the subtraction of 3x & ar in denominwor of Right hand side ring 3x22x, we get 7 1.456 X10-11 = (4x2)? (0.362)2x10-65333 upon calculastons we get { a = 4.27 x 10-4 T 30 e 2 et egner Scanned with CamScanner -

o caft equilibrium, Po = Partial pressure of D, = 24 P = 2 x 4.27X10-4 [Po= 8.54 x 10-4 bar Scanned with CamScanner
Hope you will like my answer:)