Question

Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0043 M 2N, 0(e) - 2N,(e)+0,08) Suppose a 5.0 L flask is charged under these conditions with 200 mmol of dinitrogen monoxide. How much is left 2.0 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. x I ?

Answer 1

Given:

Rate constant of a zero-order reaction = 0.0043 Ms^-1

                2N₂O (g) à 2N₂ (g) +O₂(g)

Dinitrogen monoxide: Number of moles = 200.0 mmol

Volume = 5.0 L

Time = 2.0 s

Rate of the reaction is nothing but the speed at which the chemical reaction proceeds.

It can be written two ways, either concentration of products formed or concentration of reactants consumed.

Zero-order reaction:

The rate of the reaction does not change with increase or decrease the concentration.

Example: all photochemical reaction.

The chemical reaction is written as,

                                2N₂O (g) à 2N₂ (g) +O₂(g) [zero-order reaction]

The rate of the reaction, r = -d[N₂O]/dt = k[N₂O]⁰                                                                              ---(1)

Where “-“indicate the decreases in concentration, k = rate constant

Rate constant k_o:

[N2O] is written as [A] for simplicity.

[A] = -kt + [A_o]

Where [A] = Concentration of reactant left after time “t”

[A_o] = Initial concentration of reactant

t = time

k = zero order rate constant.

Substitute the value k = 0.0043 Ms^-1, t = 2.0 s in equation [A] = -kt + [A_o]

Initial concentration of reactant [A_o] = Number of moles / Volume

[A_o] = 200.0 mmol/ 5.0L = 40 mM = 0.04 M

[A] = -kt + [A_o] becomes

[A] = - 0.0043 Ms^-1 (2.0s) + 0.04 M

[A] = -0.0086 M + 0.04 M

[A] = 0.0314 M

[A] = 0.031 M