Question

A reaction vessel is charged with 0.50 atm of A and 0.030 atm of B. Once the reaction reaches equilibrium, what is the equilibrium partial pressure of B? Kp for this reaction is 67.2A (g) ⇌ 2 B (g)

Answer 1

Answer:

1.000 atm

Explanation:

Step 1: Write the reversible reaction

A(g) ⇌ 2 B(g)

Step 2: Calculate the reaction quotient

Q = [B]²/[A]

Q = 0.030²/0.50

Q = 0.0018

since Q < Kp, the reaction will proceed to the right

Step 3: Make an ICE chart

       A(g) ⇌ 2 B(g)

I      0.50      0.030

C       -x          +2x

E    0.50-x    0.030+2x

Step 4: Find the value of "x"

We will use the definition of the pressure equilibrium constant.

Kp = 67.2 = [B]²/[A] = (0.030+2x)²/0.50-x

33.6 - 67.2x = 0.0009 + 0.12x + 4x²

4x² + 67.32x -33.5991 = 0

We solve this quadratic equation and we get x=0.485112  and x=−17.3151 (neglected).

Step 5: Find the partial pressure of B at equilibrium

pB = 0.030+2x = 0.030+2(0.485112) = 1.000 atm

Answer 2

The equilibrium partial pressure of B would be:

$1.000 atm$

Reversible Reaction

Given that,

Reaction

$A(g)$ ⇄ $2 B(g)$

Charge on the reaction vessel $= 0.50 atm$ of A

$0.030 atm$ of B

To find,

Reaction quotient first

$= B^2/A\\= (0.030)^2/0.50\\= 0.0018$

This shows that the reaction quotient is greater than Kp.

Through the reaction $A(g)$ ⇄ $2 B(g)$, ICE can be drawn

I      $0.50$      $0.030$

C    $-x$        $+2x$

E    $0.50-x$   $0.030+2x$

Now,

We will determine the value of $x$,

Kp $= 67.2$

$= [B]^2/[A]$

$= (0.030+2x)^2/0.50-x$

⇒ $33.6 - 67.2x$ $= 0.0009 + 0.12x + 4x^2$

⇒ $4x^2 + 67.32x -33.5991 = 0$

∵ $x = -17.3151$

Therefore, Equilibrium Partial Pressure

$= 0.030+2x$

$= 0.030+2(0.485112)$

= $1.000$ atm

Learn more about "Partial Pressure" here:

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