An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)-- 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?

Question

An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)--> 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?

Serenity Allen · Accepted Answer

The answer is #3.3M#.  The best way to approach this problem is by using the ICE table method (more here: https://tutor.hix.ai). We know that, initially, a concentration of #NO_2#, #A#, was placed in the flask, which means that the initial concentrations of #NO# and #O_2# were zero. We must use the reverse reaction, since no amounts of #NO# and #O_2# are present at the start; this means that the equlibrium constant will be equal to #K_(reverse) = 1/K_(fo rward) = 1/24.0 = 0.0417# SInce the equilibrium constant for the reverse reaction is smaller than 1, the final mixture will contain mostly reactants, which means that the concentration of #NO_2# at equlibrium (and of course, before the reaction) must be greater than those of #NO# and #O_2#. .....#2NO_(2(g)) rightleftharpoons 2NO_((g)) + O_(2(g))#
I.........A.................0...................0
C......-2x...............+2x................+x
E.......(A-2x)...........2x.................x Since we know that #NO#'s equilibrium concentration is 0.800 M, we get #2x = 0.800 -> x= 0.800/2 = 0.400M# The expression for the equilibrium constant is #K_(eq) = ([NO]^2 * [O_2])/([NO_2]^2) = (0.800^2 * 0.400)/(A-0.800)^2 = 0.0417# #(A-0.800)^2 = (0.800^2 * 0.400)/0.0417 = 6.14# Solving for #A# will produce two values, one negative and one positive; since #A# represents concentration, which cannot be negative, the only value for #A# will be 3.3. Therefore, #NO_2#'s initial concentration is #3.3M#. Notice that the prediction about #NO_2#'s concentration at equilibrium was accurate; at equilibrium, the species have the following concentrations: #[NO_2] = 3.3 - 0.8 = 2.5M#
#[NO] = 0.8M#
#[O_2] = 0.4M#

Luna Chen · Answer

# 2"NO"_2"# = "NO" + "O"_2

E/mol·L⁻¹: 2#x#; #x#; #C - 2x# I/mol·L⁻¹: 0; 0; #C# C/mol·L⁻¹: +2#x#; +#x#; -#2x#

We are aware that 2x = 0.800# = #["NO"]_"eq"

Thus, #x = 0.400#

"eq" #[O_2] = x = 0.400#

"eq" #[NO_2] = C – 2x = C -0.800#

Step 3: Compose the expression #K_"eq"#.

["NO"_2]^2/(["NO"^2]["O"_2]) = #K_"eq" = 24.0#

Step 4: Enter these values into the expression #K_"eq"#.

(C - 0.800)^2/(0.800^2 × 0.400) = 24.0# #K_"eq" = ["NO"_2]^2/(["NO"^2]["O"_2])

Step 5: Determine #C#.

#(C-0.800)^2 = 24.0 by 0.640 by 0.400 = 0.800^2 by 0.400) = 6.144#

6.144# is #C^2 – 1.600C + 0.640.

5.504 - 1.600C - #C^2 = 0#

#C equals 3.279#

The NO2 concentration was 3.279 mol/L at the beginning.

Verify:

"eq" #["NO"] = 0.800#

"eq" #[O_2] = x = 0.400#

C -2x = 3.279 – 0.800 = 2.479# #[NO_2]_"eq"

#K_"eq" = ["NO"_2]^2/(["NO"^2]["O"_2]) = 2.479^2/(0.800^2 × 0.400) = 24.0#

Mateo Aguilar · Answer

undefined Initial concentration of NO2: 0.200 M.

An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)--&gt; 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?

An amount of NO2 was initially put into a 5.0L flask. When equilibrium was attained according to the equation: 2NO (g) + O2 (g)--> 2NO2 (g), the concentration of NO was 0.800M. If Keq for this system is 24.0, what was the initial concentration of the NO2?