A sample of nitrogen dioxide is placed in a container at 250°C and 1 atm, where it decomposes according to the following equation: 2NO2(g) ↔ 2NO(g) + O2(g)The initial concentration of NO2 is 0.10 M, and the equilibrium constant (Kc) at this temperature is 4.0 x 10^-3. If the volume of the container is suddenly decreased to one-half of its original volume, causing the pressure to double, what will happen to the amount of NO and O2 present at equilibrium? Calculate the new equilibrium concentrations of NO and O2.

Question

A sample of nitrogen dioxide is placed in a container at 250°C and 1 atm, where it decomposes according to the following equation: 2NO2(g) ↔ 2NO(g) + O2(g)The initial concentration of NO2 is 0.10 M, and the equilibrium constant (Kc) at this temperature is 4.0 x 10^-3. If the volume of the container is suddenly decreased to one-half of its original volume, causing the pressure to double, what will happen t

Answer 1

First, let's find the initial equilibrium concentrations of NO and O2. We can set up an ICE  Initial, Change, Equilibrium  table to help us with this:``          2NO2 g       2NO g    +   O2 g Initial:   0.10 M        0 M          0 MChange:   -2x           +2x          +xEquilibrium: 0.10-2x     2x           x``Now we can set up the equilibrium expression using the given Kc value:Kc = [NO]^2[O2] / [NO2]^2 = 4.0 x 10^-3Plugging in the equilibrium concentrations from the ICE table: 4.0 x 10^-3  =  2x ^2 x  /  0.10 - 2x ^2Solving for x, we get:x  0.0146Now we can find the initial equilibrium concentrations:[NO2] = 0.10 - 2x  0.071[NO] = 2x  0.029[O2] = x  0.0146Now let's consider the effect of decreasing the volume to half of its original volume. The initial concentrations of all species will double:[NO2] = 2 0.071  = 0.142 M[NO] = 2 0.029  = 0.058 M[O2] = 2 0.0146  = 0.0292 MNow we can set up a new ICE table for the new conditions:``          2NO2 g       2NO g    +   O2 g Initial:   0.142 M       0.058 M      0.0292 MChange:   -2y           +2y          +yEquilibrium: 0.142-2y    0.058+2y     0.0292+y``We can use the same Kc value since the temperature has not changed:Kc = [NO]^2[O2] / [NO2]^2 = 4.0 x 10^-3Plugging in the new equilibrium concentrations from the ICE table: 4.0 x 10^-3  =  0.058+2y ^2 0.0292+y  /  0.142 - 2y ^2Solving for y, we get:y  0.0058Now we can find the new equilibrium concentrations:[NO2] = 0.142 - 2y  0.130[NO] = 0.058 + 2y  0.070[O2] = 0.0292 + y  0.035So, the new equilibrium concentrations after decreasing the volume are approximately:[NO2]  0.130 M[NO]  0.070 M[O2]  0.035 MAs a result, the amount of NO and O2 present at equilibrium has increased due to the decrease in volume and the increase in pressure.