A 1-L container originally holds 0.4 mol of N_2, 0.1 mol of O_2, and 0.08 mole o

Question

A 1-L container originally holds 0.4 mol of N_2, 0.1 mol of O_2, and 0.08 mole of NO. If the volume of the container holding the equilibrium mixture of N_2, O_2, and NO is decreased to 0.5 L without changing the quantities of the gases present, how will their concentrations change? The concentrations of N_2 and O_2 will increase; and the concentration of NO will decrease. The concentrations of N_2, O_2, and NO will decrease. The concentration of NO will increase; the concentrations of N_2 and O_2 will decrease. There will be no change in the concentrations of N_2, O_2, and NO. The concentrations of N_2, O_2, and NO will increase.

Explanation / Answer

Q8

[N2] = 0.4 M

[O2] = 0.1

[NO] = 0.08

V decreases by half, so concentration must increaase by double i.e. 1/2 --> 2/1

[N2] = 0.8 M

[O2] = 0.2

[NO] = 0.16

Then

note that there is no information on equilibrium... so assume the equilibrium is favoured for none

since

N2 + O2 = 2NO, 2 mol of products in gas = 2 mol of gas in reactants

so...

E is the correct answer,

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