Question

he equilibrium constant (KP) is 0.21 at a particular temperature for the reaction:

N2O4(g) <=> 2NO2(g)

Given the following sets of initial conditions, what is the net change that must occur for the reaction to reach equilibrium? Does the reaction shift left to reach equilibrium, does the reaction shift right to reach equilibrium or is the reaction at equilibrium at these initial concentrations so no net change will occur?

Options: Equilibrium, Left, or Right

a. PNO2 = 0.225 atm, PN2O4 = 0.134 atm

b. PNO2 = 0.086 atm, PN2O4 = 0.064 atm

c. PNO2 = 0.130 atm, PN2O4 = 0.142 atm

d. PNO2 = 0.173 atm, PN2O4 = 0.142 atm

e. PNO2 = 0.151 atm, PN2O4 = 0.062 atm

Answer 1

Explanation:

We will use the following equation,  

     $Q_{p} = \frac{(P^{2}_{NO_{2}}{P_{N_{2}O_{4}}$

(a) Now, we will calculate the value of $Q_{p}$ as follows.

          $Q_{p} = \frac{(0.225)^{2}}{0.134}$

                      = 0.377

As, the $K_{p}$ is 0.21 (which is smaller than the value) this means that equilibrium will shift to the left .

(b)   $Q_{p} = \frac{(0.086)^{2}}{0.064}$

                  = 0.1155

Here, $Q_{p}$ is smaller than $k_{p}$ which means that the equilibrium will shift to the right.

(c)   $Q_{p} = \frac{(0.13)^{2}}{0.142}$

                   = 0.119

Here, $Q_{p}$ is smaller than $k_{p}$ which means that the equilibrium will shift to the right .

(d)    $Q_{p} = \frac{(0.173)^{2}}{0.142}$

                  = 0.21  

Since, here the value of $Q_{p}$ comes out to be equal to the given value. Therefore, this reaction is at equilibrium.

(e)  $Q_{p} = \frac{(0.151)^{2}}{0.062}$

                  = 0.36

Here, $Q_{p}$ is higher than $k_{p}$ the reaction will shift to the left.

Answer 2

The reaction shifts to the left or right depending on the values of the reaction quotient (Q).

The reaction is shown as, N2O4(g) <=> 2NO2(g) and we are told that Kp =  0.21.

In each of the following conditions, we need to calculate Q in order to make the required decision. Recall that Q = [NO2]^2/[N2O4]

For (a);

PNO2 = 0.225 atm, PN2O4 = 0.134 atm

Q = [0.225]^2/[0.134] = 0.38

In this case K < Q, the reaction must shift to the left to attain equilibrium

For (b);

PNO2 = 0.086 atm, PN2O4 = 0.064 atm

Q = [0.086]^2/[0.064]

Q = 0.115

Since K > Q, the reaction must shift in the forward direction

For (c);

PNO2 = 0.130 atm, PN2O4 = 0.142 atm

Q = [0.130]^2/[0.142]

Q = 0.119

Since K > Q, the reaction must shift in the forward direction

For (d);

PNO2 = 0.173 atm, PN2O4 = 0.142 atm

Q = [0.173]^2/[0.142]

Q = 0.21

Since K = Q, the reaction has attained equilibrium

For (e);

PNO2 = 0.151 atm, PN2O4 = 0.062 atm

Q = [0.151]^2/[0.062]

Q= 0.37

In this case K < Q, the reaction must shift to the left to attain equilibrium

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