Question

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

Short answer

For part a, the equilibrium partial pressures for N2O4 and NO2 are 3.97 atm and 1.06 atm, respectively. For part b, they are 1.097 atm and 6.806 atm, respectively. Comparing the results, the ratios of equilibrium partial pressures for both cases are approximately the same (0.267), indicating that it doesn't matter from which direction the equilibrium position is reached, as long as the reaction is at equilibrium and has the same Kp value.

Step-by-step solution

Write the expression for Kp

For the given reaction, the expression for Kp can be written as: \[ K_{\mathrm{p}} = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} \] #a#

Set up the ICE table for part a

Let x be the change in concentration of N2O4 and NO2 during the reaction. The ICE (Initial, Change, Equilibrium) table for part a will look like this: | | N2O4 | NO2 | |------|--------------|--------------| |Initial | 4.5 | 0 | |Change | -x | + 2x | |Equilibrium | 4.5 - x | 2x |

Substitute the given Kp value and set up the equation

Now, we plug in the values from the ICE table into the Kp expression: \[ 0.25 = \frac{(2x)^2}{(4.5 - x)} \]

Solve the equation for x

Solve the quadratic equation to find the value of x to get the equilibrium partial pressures: \[ 0.25(4.5 - x) = 4x^2 \Rightarrow x^2 = 0.28125\] \[ x = \pm 0.53\] Since the concentration cannot be negative, x = 0.53.

Calculate the equilibrium partial pressures for part a

Now, substitute the value of x in the expressions for equilibrium partial pressures of N2O4 and NO2: \[ [\mathrm{N}_{2}\mathrm{O}_{4}] = 4.5 - x = 3.97 \,atm \] \[ [\mathrm{NO}_{2}] = 2x = 1.06\, atm \] The equilibrium partial pressures of N2O4 and NO2 for part a are 3.97 atm and 1.06 atm, respectively. #b#

Set up the ICE table for part b

For part b, the initial pressure of NO2 is given. The ICE table for part b will look like this: | | N2O4 | NO2 | |------|--------------|--------------| |Initial | 0 | 9 | |Change | +x/2 | -x | |Equilibrium | x/2 | 9 - x |

Substitute the values and set up the equation

Plug in the values from the ICE table into the Kp expression: \[ 0.25 = \frac{[(9 - x)]^2}{(x/2)} \]

Solve the equation for x

Solve the quadratic equation to find the value of x and get the equilibrium partial pressures: \[ 0.25\frac{x}{2} = (9 - x)^2 \Rightarrow x^2 = 4.815625\] \[ x = \pm 2.194\] Since the concentration cannot be negative, x = 2.194.

Calculate the equilibrium partial pressures for part b

Now, substitute the value of x in the expressions for equilibrium partial pressures of N2O4 and NO2: \[ [\mathrm{N}_{2}\mathrm{O}_{4}] = x/2 = 1.097\,atm \] \[ [\mathrm{NO}_{2}] = 9 - x = 6.806\,atm \] The equilibrium partial pressures of N2O4 and NO2 for part b are 1.097 atm and 6.806 atm, respectively. #c#

Compare the results

Comparing the results obtained in part a and part b, we can see that the ratios of equilibrium partial pressures for both cases is the same. Part a: \[ \frac{[\mathrm{NO}_{2}]}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = \frac{1.06}{3.97} \approx 0.267 \] Part b: \[ \frac{[\mathrm{NO}_{2}]}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = \frac{6.806}{1.097} \approx 0.267 \] This indicates that it doesn't matter from which direction the equilibrium position is reached, as long as the reaction is at equilibrium and has the same Kp value.