Question

Consider the following reaction: A(g)⇌2B(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.

1.PA, PB, Kp= 1.4

2.PA, PB, Kp= 1.8×10^−4

3.PA, PB, Kp= 2.0×10^5

Answer 1

first to draw ICE table.

then write the equilibrium constant equation kp.

then using the given condition to find P_{​​​​​​A} and P_{​​​​​​B}

A Soln 2B initial ch x 1-2x Withing the equlibrium.cometent sexu Ko - trobe . ti-za) ( int given upeliu (-28) 1.4 0 + 4x 4x al. 4x 4x - 5.40 + 1 =0 x=5.4 + J (5-4) 2-4X1XY 1 2x4 x=002215 or. (1-128 this x=0-2215 atm. Po = x= 10. 2215 atm. Pa = 1-280-2215 +0.557|atm. not possible. © Kp=118810-Y (1-2x) = 108*10-4 31+ 4x²_4X = 108 x 16-7X 4x² - 4x + 1 = 0 x = 4 I 54%=4x4 - - 2x 4 PA = 10.5 atm. PA = 1-2x0,5=10 atm. 0.5 -

2xiOS Date --I-I I Kp = 20x105 (1-200² - x It Y -40 = 2x10 x. yx?_ 2 x 10*x+d=0 ex10s + (2+105)² - 4X4XL 2x4 *=581006 Ø 50000 punet persible PA =T5 x 10-6 latm. Po = 1-2x5x10-6 = 1 atm.