An exothermic reaction was performed at 298 Kelevin and in a 1 L canister. A(s)

Question

An exothermic reaction was performed at 298 Kelevin and in a 1 L canister. A(s) + 2B (g) <--> C(g)+ D(g). Kp = 4.68*10^-4

1. If 20.0 grams of molecule A, molar mass 100.4 grams per mole, is placed in a canister and filled with 0.4 atm of molecule B, what is the pressure of B at equilibrium pressure? (Ans: 0.387 atm)

2. If 20.0 grams of molecule A, molar mass 100.4 grams per mole, is placed in a canister and filled with 0.4 atm of molecule B, how many grams of A had to react to get to equilibrium? (Ans. 19.5 grams)

Explanation / Answer

For the reaction,

Kp = [C][D]/[A][B]^2

1. moles A = 20/100.4 = 0.2 mol

Initial pressure of A = 0.2 x 0.08205 x 298/1 = 5 atm

ICE chart

                     A   +   2B <===>   C    +   D

I                    5         0.4               -          -           

C                 -x         -2x               +x       +x

E                (5-x)   (0.4-2x)           x         x

So,

4.68 x 10^-4 = x^2/(5-x)(0.4-2x)^2

0.16 - 1.6x + 4x^2

0.8 - 8x + 20x^2 - 0.16x + 1.6x^2 - 4x^3

-1.87 x 10^-3x^3 - 0.99x^2 - 3.82 x 10^-3x + 3.74 x 10^-4 = 0

x = 0.01 atm

So,

equilibrium pressure of B = 0.4 - 2 x 0.01 = 0.38 atm

2. Pressure of A at equilibrium = 5 - 0.01 = 4.99 atm

change = 0.01 atm

grams change = 0.01 x 1 x 100.4/0.08205 x 298 = 0.041 g

grams reacted = 20 -0.04 = 19.9 g

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