Dinitrogen tetroxide partially decomposes according to the following equilibrium
ID: 951573 • Letter: D
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 10-30.154
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 FalseConsider 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 ________.
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