Question

Question 4: (1 point) A container at 35.0 °C initially contains AB (g) in equilibrium with AB (1) AB (1) AB (g) A catalyst (of negligible voluite) is added to the container such that AB (g) quickly decomposes and reaches equilibrium al 35.0 PC accorcling to the reaction: AB(y) A(g) +B(g) K;=0.700 at 35.0°C If the initial tolal pressure of the container (with AB only) is 0.901 bar, what is the total pressure (in bar) of the container (still containing liquid AB) al cquilibrium? Assuinc A and B do not condense and are insoluble in liquid AB. Round your answer to three significant figures. bar

Answer 1

AB (g) $\to$ A (g) + B (g)

Initial pressure is due to AB (g) only.

Now ICE table is

	PAB , bar	PA , bar	PB , bar
I	0.901	0	0
C	-x	+x	+x
E	(0.901-x)	x	x

Now, K_p =

PA x Pв РАВ

So, K_p =
_{.CXC (0.901 - .)}

Or, 0.700 =
_{(0.901 -r)}

Or, 0.700×( 0.901 - x) = x²

Or, 0.6307 - 0.700x = x²

Or , x² + 0.700 x - 0.6307 = 0

a = 1; b = 0.700 , c = - 0.6307

Solving quadratic equation.

X =

-0.700 + (0.700)2 – 4*1*(-0.6307) 2 x 1

Or, x =

-0.7 +1.736

Only positive value is taken ( pressure can not be negative)

Or, x = 0.5178  

Now, pressure of A = B = 0.5178 bar

Pressure of AB = (0.901 - 0.5178) = 0.3831 bar

Total pressure = P_A+ P_B + P_AB

= 0.5178 + 0.5178 + 0.3831

= 1.4187 bar

= 1.42 atm (three significant figure).

( calculated Kp = (0.5178)²/(0.3831) = 0.699 , hence answer is valid) .