Question

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the equilibrium pressure of NO2 if Kp for the reaction at this temperature is 0.0118. NO(g) SO3(g) NO2(g) SO2(g)

Answer 1

Explanation:

Reaction equation for this reaction is as follows.

     $NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

It is given that $K_{p}$ = 0.0118.

According to the ICE table,

              $NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

Initial:           0.86      0.86               0            0

Change:          -x          -x                 +x           +x

Equilibrium:  0.86 - x   0.86 - x         x           x

Hence, value of $K_{p}$ will be calculated as follows.

           $K_{p} = \frac{P_{NO_{2}} \times P_{SO_{3}}}{P_{NO} \times P_{SO_{3}}}$

         0.0118 = $\frac{x \times x}{(0.86 - x)^{2}}$

             x = 0.084 atm

Thus, we can conclude that $P_{NO_{2}}$ is 0.084 atm.

Answer 2

0.084 atm is the equilibrium pressure of NO₂, if Kp for the reaction at this temperature is 0.0118.

How we calculate equilibrium constant from partial pressure (Kp)?

Equilibrium constant from partial pressure (Kp) can be calculated as:

Kp = [product of the concentration of products] / [product of the concentration of reactants]

Given chemical reaction can be expressed in terms of initial, change, equilibrium as follow:

                         NO(g) + SO₃(g) → NO₂(g) SO₂(g)

Initial:                 0.86       0.86        0          0

Change:              -x             -x         +x        +x

Equilibrium:    0.86-x      0.86-x      x           x

From the above data expression for Kp is written as:

Kp = [NO₂] [SO₂] / [NO] [SO₃]

Kp = x.x / (0.86-x)²

In the question given value of Kp = 0.0118

Putting this value in the above equation, we get

0.0118 = x² / (0.86-x)²

On the solving the quadratic equation we get the value of x is 0.084 atm.

Hence, the equilibrium pressure of NO₂ is 0.084 atm.

To know more about Kp, visit the below link:

https://brainly.com/question/25651917