For part A I got 1.58 M CH 15 HMWK Exercise 15.98 Exercise 15.98 When N2Os (g) i

Question

For part A I got 1.58 M CH 15 HMWK Exercise 15.98 Exercise 15.98 When N2Os (g) is heated it dissociates into N20,(g) and O2(g) according to the following reaction: N20's(g) N203 (g)+O2 (g) 7.75 at a given temperature. The N203(g) dissociates to give N20(g) and O2(g) according the following reaction N203 (g) N20(g) t O2 (g) Ke 400 at the same temperature. When 4.00 mol of N2Os(g) is heated in a 1.00-L reaction vessel to this temperature, the concentration of O2 (g) at equilibrium is 4.50 mol/L. EE O Type here to search You completed this assignment. Part A Find the concentration of N20s in the equilibrium system Express your answer using two decimal places. N20 1.58 M Submit My Answers Give Up Answer Requested Part B Find the concentration of N20 in the equilibrium system Express your answer using two decimal places. (N20) 2.25 Submit Incorrect: Try Again: 3 attempts remaining

Explanation / Answer

N2O5 = N2O3 + O2

K = 7.75

N2O3 = N2O + O2

K = 4

Overall equation.

N2O3 + N2O5 = N2O3 + O2+ N2O + O2

cnacel common terms

N2O5 = N2O + 2O2

Overall K --> K1*K2 = 7.75*4 = 31

now

K = [N2O][O2]^2 / [N2O5]

initially:

[N2O5] = 4/1 = 4 M

[O2] = 0

[N2O] = 0

in equilibrium

[N2O5] =4 - x

[O2] = 0 + 2x

[N2O] = 0 + x

and we know that

[O2] = 4.5

so

[O2] = 0 + 2x =4.5

x = 4.5/2 = 2.25

substitute:

[N2O5] =4 - x = 4-2.25 = 1.75 M

[O2] = 0 + 2x = 2*2.25 = 4.5 M

[N2O] = 0 + x = 0 + 2.25= 2.25 M

B)

[N2O] = 2.25 M

C)

[N2O3] = from initial equilibrium

N2O5 = N2O3 + O2

K = [O2][N2O3] / [N2O5]

4 = 4.5 * X /1.75

X = 4/4.5 * 1.75

X = [N2O3] = 1.55555 M

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