Consider the reaction of ammonium ions (NH4+)and nitrite ions (NO2-) NH4+(aq)+ N

Question

Consider the reaction of ammonium ions (NH4+)and nitrite ions (NO2-)

NH4+(aq)+ NO2-(aq)--> N2(g) + 2H2O (l)

Solutions containing NH4+ andNO2- were mixed in various quantities and thefollowing rate data at a constant temperature temperature wereobtained:

Determination
1
2
3

INITIAL [NH4+] CONCENTRATION MOL/L
0.150
0.150
0.300

Initial [NO2-] CONCENTRATION MOL/L
7.50 * 10^-3
1.50 * 10^-2
1.50 * 10^-2

INITIAL RATE FOR FORMATION OF N2, MOL/L * S
3.04 * 10^-7
6.08 * 10^-7
1.22 * 10^-6

A. USE THE METHOD OF INITIAL RATES TO FIND THE ORDER OF THE REACTION WITH RESPECT TO NH4+.
B. USE THE METHOD OF INITIAL TO FIND THE ORDER OF THE REACTION WITH RESPECT TO NO2-.
C. CALCULATE THE RATE CONSTANT, K, FOR THE REACTION OF NH4+ AND NO2-.
D. WRITE THE RATE EQUATION FOR THE REACTION OF NH4+ AND NO2-.

Explanation / Answer

a )the first thing we should do is find the order of each individual reactant. we use the equation: Rate1/Rate2 = ( [ClO2 1] / [ClO2 2] )^a * ( [OH- 1] / [OH- 2] )^b. where the 1's and 2's represent one of the runs, and a and b represent the order. we round to the nearest whole number. it doesn't matter which u use, as long as ur consistent. Obviously, we have to find 2 runs in which one substance's concentration doesn't change , since we can only have 1 variable, and u'll get 1^a = 1 between runs 2 and 3 , the ClO2 doesn't change. so, first we'll find the rate order for OH-. when both have the same power of 10 attached, u can ignore it in the equation, since they cancel. (9.34 x 10^ -1 / 1.87 ) = (1.30 / 2.60)^b {10^ -4 / 10^ -3 = 10^ -1} 0.5 = 0.5^b b=1 so now we have that OH- is first order. now, we use runs 1 and 2 (or 1 and 3) to find the order of ClO2 since we need the concentration to change. 2.33 / 9.34 = (1.25 / 2.5)^a 0.25 = 0.5^a a=2. so the rate equation is Rate = k[OH-] [ClO2]^2. now: b ) to find the rate constant, plug in the numbers for one of the runs into the rate law. lets use run 1. Rate = k[OH-] [ClO2]^2 2.33 x 10^-4 = 1.30 x 10^ -2 x (1.25 x 10^ -2)^2 x k. 2.33 x 10^ -4 = 1.5625 x 10^-6 x k k = 149 or 1.49 x 10^2. rate law is now Rate = 149[OH-] [ClO2]^2 c) to find the reaction rate for another circumstance, just plug in all the numbers. Rate = 149 x 5.35 x 10^-2 x (8.25 x 10^-3)^2 rate = 5.42 x 10^-5

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