Question

Consider the following reaction: A (g) = 2 B (g) Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.

Answer 1

Initially 0.0 A²2B - latm xatm (1-2x) alm where x is partial pressure of A at cowlibrium Ayhan At equilibrium Part A. Kp = (PB) ² = (1-2x)² 1-4 1 + 4x2 - 4x = 1.4x = 4X-4X1 - 4x²_s.txt =0 which gives x= 1-128 and 0.22 · We will conside x = 0.22 because x: 1.128 gives negative preso of B - [PA = 0.22 cm Po= 0.66 | Part B - Kp= (P2)? - 144x² AX 10-42 = 1+4x?- 4x − 4x2 - 4x + 20 which give x = +% so final (PA = 0.Salm PB= 0.0atm] Partec Kp = Pig)? 1+4x24u 1.2.105€ 1+4x² - 4x *****ZAMBARAKELO 1-2x105(1-2x) This large value can be obtained only when value of a is very 12 x105 = small in comparison to 1 so we assume x = 8-33 xlot atm 1-2x n 1 x PR = 1 atm cs Scanned with Pan O ato CamScanner