A(g) + 2B(g) → C(g) + D(g)

If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

screening

explanation:

a sequence is a function whose domain is the set of natural numbers.

natural numbers are all the negative and positive integers