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

Explanation:

We will use the following equation,  

(a) Now, we will calculate the value of as follows.

                      = 0.377

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

(b)  

                  = 0.1155

Here, is smaller than which means that the equilibrium will shift to the right.

(c)  

                   = 0.119

Here, is smaller than which means that the equilibrium will shift to the right .

(d)    

                  = 0.21  

Since, here the value of comes out to be equal to the given value. Therefore, this reaction is at equilibrium.

(e)  

                  = 0.36

Here, is higher than the reaction will shift to the left.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%