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Brown's Chemistry: The Central Science, 15.8 Exercise 1

Posted by Dr. Fred Zhang | Jan 16, 2019 7:00:00 PM

Teaching Explanations

 

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This posts contains a Teaching Explanation.

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Exercise: 15.8 Practice Exercise 1

Question: … When 9.2g of frozen $N_2O_4$ is added to a .50L reaction vessel … [What is the value of $K_c$]

 

Part 1:  Approaching the Problem

The question is asking for an equilibrium constant ($K_c$).  We want to know $K_c$.  

Generally, we can know the equilibrium constant ONLY IF we can figure out the equilibrium concentrations of the species (nitrous oxide and dinitrogen tetraoxide):

$$K_c = [NO_2]^2/[N_2O_4]$$

Thus, the entire game to figuring out the equilibrium constant here is to figure out the equilibrium concentrations.

We are already given that in equilibrium, the concentration of $[N_2O_4]$=.057 molar.  So we have half the puzzle:

$$K_c = [NO_2]^2/.057$$

The other half of the puzzle if figuring out the equilibrium concentration $[NO_2]$.  Sadly, the question doesn’t just give us this.  But we have a piece of information nearly as good, which is the starting (initial) amount of $[N_2O_4]$.  Because we know the reaction equation, the key now is to go from initial amount of $[N_2O_4]$ to the final (equilibrium) concentration $[NO_2]$.  

 

Part 2: Converting Grams to Molar

We are given that the reaction started out with 9.2g of $N_2O_4$ in a 0.50L reaction vessel.  For equilibrium calculations, we generally want to know concentrations of types molecules, instead of actual mass or volume.  

We apply stoichiometry here and convert grams per liter to molarity using molar mass.  We use the periodic table to look up the molar mass of $N_2O_4$ is 92.01 grams per mole.

We get that:

$$(9.2g N_2O_4)/(0.50L) *(1 mol)/(92.01 g N_2O_4) = (0.100mol)/L = 0.200 molar$$

Thus the initial concentration of $N_2O_4$ is 0.200 molar, and written as [$N_2O_4$]=.200

 

Part 3: Running the Reaction

Now that we know the starting concentration, we want to get to final concentrations.  The algebraic equation that links the two is the equation of reaction:

$$N_2O_4 (g) ↔ 2 NO_2 (g)$$

This means that for every molecule of $N_2O_4$ we get two molecules of $NO_2$.  As the reaction goes forward, when $N_2O_4$ decreases by $x$ molar, $NO_2$ increases by $2x$ molar.  

The concentration table is then:

 

$N_2O_4 (g)$

$2 NO_2 (g)$

Initial Concentration (M)

0.200

0

Change in Concentration (M)

-x

+2x

Equilibrium Concentration (M)

0.200-x

2x

 

 

Part 4: Calculating the Equilibrium 

We are given that the equilibrium concentration of [$N_2O_4$]=.057 molar.  The concentration table above gives the equilibrium concentration of [$N_2O_4$]=0.200-x, so we just equate the two and solve for x.

0.200-x = 0.057

x = .143

Now that we know x,

2x = .268

Or that in equilibrium, $[NO_2]=.268$

To calculate the equilibrium constant Kc, we plug in the information above:

$$K_c = [NO_2]^2/[N_2O_4]=.268^2/.057 = 1.43$$

Therefore, the right answer is d) 1.4



Video Solution

Brown Ch 15.7 Excercise 1

 

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Dr. Fred Zhang
About the Author

Fred is co-founder of PrepScholar. He scored a perfect score on the SAT and is passionate about sharing information with aspiring students. Fred graduated from Harvard University with a Bachelor's in Mathematics and a PhD in Economics.



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