Consider the hypothetical reaction A (g) 2B(g). A flask is charged with 0.74 atm
ID: 987386 • Letter: C
Question
Consider the hypothetical reaction A (g) 2B(g). A flask is charged with 0.74 atm of pure A, after which it is allowed to reach equilibrium at 0 degree C. At equilibrium the partial pressure of A is 0.35 atm. What is the total pressure in the flask at equilibrium? Express your answer using two significant figures. What is the value of K_p? Express your answer using two significant figures. What could we do to maximize the yield of B? Doing the reaction in a larger flask maximizes the yield of B. Doing the reaction in a smaller flask maximizes the yield of B.Explanation / Answer
A(g) <------> 2B(g)
initally ...
0.74 <----> 0
at equilibrium...
0.74-x <-----> 2x
so equilibrium partial pressure of A = 0.74-x = 0.35
x = 0.74-0.35 = 0.39 atm
equilibrium partial pressure of B = 2x = 2 X 0.39 = 0.78 atm
(a)total pressure of flask at equilibrium = equilibrium partial pressure of A + equilibrium partial pressure of B = 0.39 + 0.78 = 1.17 atm
(b)Kp = P^2(B)/P(A) = (1.17)^2/0.35 = 3.9111