Consider the reaction of ammonium ions (NH 4 + )and nitrite ions (NO 2 - ) NH 4

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 intial [NH4+], M
initial [NO2-], M
+[N2]/t = initial rate of N2, M*s 1 0.150 7.50*10-4
3.04*10-7
2 0.150 1.50*10-3
6.08*10-7
3 0.300 1.50*10-3
1.22*10-6

The question is, for determination 3, the initial rate of formationof N2 is given as 1.22*10-6 M*s, what is therate of formaiton of H2O?

After finding the rate = k[a]x[b]y

I found that X = 1, Y = 1, and K = 0.0027

Then I'm stuck, from the equation, can I just say that because 1mole of ammonium ion creates 1 mole of N2, therefore therate of formation of H2O is half of what N2gives me?

Determination intial [NH4+], M
initial [NO2-], M
+[N2]/t = initial rate of N2, M*s 1 0.150 7.50*10-4
3.04*10-7
2 0.150 1.50*10-3
6.08*10-7
3 0.300 1.50*10-3
1.22*10-6

Explanation / Answer

The rate of formation of H2O will be2.44x10-6. or twice as fast as N2. If you look at the stoichiometric coefficients in your balancedequation, you will see that for every N2 formed thereare 2 H2O formed. So in terms of your question, if N2 forms at the rate of1.22x10-6, H2O has to form twice as fast, or2.44x10-6. Clear as mud? Hope this helps, Steve

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