Dinitrogen tetroxide partially decomposes according to the following equilibrium

Question

Dinitrogen tetroxide partially decomposes according to the following equilibrium:

N2O4 (g)   2NO2 (g)

A 1.000-L flask is charged with 8.00 10-3 mol of N2O4. At equilibrium, 5.04 10-3 mol of N2O4 remains. Keq for this reaction is ________.

Dinitrogen tetroxide partially decomposes according to the following equilibrium:

(g)    (g)

A 1.000-L flask is charged with 8.00  10-3 mol of . At equilibrium, 5.04  10-3 mol of  remains.  for this reaction is ________.

0.154

The effect of a catalyst on a chemical reaction is to react with product, effectively removing it and shifting the equilibrium to the right.

The effect of a catalyst on a chemical reaction is to react with product, effectively removing it and shifting the equilibrium to the right.

0.181 2.37 10-5 0.212 6.94 10-3

0.154

The effect of a catalyst on a chemical reaction is to react with product, effectively removing it and shifting the equilibrium to the right.

The effect of a catalyst on a chemical reaction is to react with product, effectively removing it and shifting the equilibrium to the right.

True False

Consider the following reaction at equilibrium:

2CO2 (g)   2CO (g) + O2 (g) H° = -514 kJ

Le Châtelier's principle predicts that removing O2(g) to the reaction container will ________.

Consider the following reaction at equilibrium:

2CO2 (g)   2CO (g) + O2 (g) H° = -514 kJ

Le Châtelier's principle predicts that removing O2(g) to the reaction container will ________.

increase the value of the equilibrium constant increase the partial pressure of CO2 decrease the partial pressure of CO2 decrease the partial pressure of CO decrease the value of the equilibrium constant

Explanation / Answer

a)

First, get K

N2O4 (g) 2NO2 (g)

K = [NO2]^2 / [N2O4]

initially

[N2O4] = 8*10^-3 = 0.008

[NO2] = 0

in equilibrium

[N2O4] = 0.008 - x

[NO2] = 0 + 2x

and we know that:

[N2O4] = 5.04*10^-3 = 0.00504

[N2O] = 0.008 - x =0.00504

solve for x

x = 0.008-0.00504 = 0.00296

so

[NO2] = 0 + 2x = 2*0.00296 = 0.00592

Keq = (0.00592^2)/0.00504

Keq = 0.0069536 = 6.95*10^-3

Q2

False, addition of catalyst will only make reaction faster, but will not make any shift in the equilibrium ratios

Q3.

2CO2 (g) <=> 2CO (g) + O2 (g) H° = -514 kJ

If we add O2, a product, then.. .according to Le Chatelier...

There will be a shift to the left, since there is exces sproducts

so expect more CO2 formation

expect CO decrease

TRUE statements:

increase the partial pressure of CO2

decrease the partial pressure of CO

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