4. The reaction O2(g) N2(g) 2NO(g) was prepared with the following initial conce

Question

4. The reaction O2(g) N2(g) 2NO(g) was prepared with the following initial concentrations: [O2 0.01 M [N2] 0.01 M and [NO) a 0 M. At equilibrium, [NO] was measured and found to 0.00020 a. set-up an ICE table and calculate the equilibrium concentration for the two reactants O2 and N2. b. Calculate Kc. 5. The same reaction is now run with the following initial concentrations: O2(g) N2 (g) 2NO(g) [o2] 0.1 M [N2] 0.0001 M and [NO) o M Because the concentration of oxygen is one thousand times as concentrated as that of nitrogen, the reaction is driven to the right effectively using up all of the N2. a. Using an ICE table, calculate all the equilibrium concentrations for all species in this reaction. b. Do the results of your ICE show that it is a correct assumption that almost all of the N2 placed in the vessel will be used up under these conditions?

Explanation / Answer

(4)

O2(g)+N2(g)----->2NO(g)

Take 1 lit of solution

So 0.01 M =0.01 mole/lit

So in 1 lit 0.01 mole

According to stoichiometry

1 mole of O2 and 1 mole of N2 produce 2 mole of NO.

So at equlibrium NO moles given =0.0002 moles

According to stoichiometry at equlibrium 0.0001 moles of N2 and O2 produce 0.0002 moles NO

So change= 0.1-0.0001=0.0999

So equlibrium constant

Kc = [NO]2/[N2]*[O2] = (0.0002)2 / (0.0001)*(0.0001) =4

(5)

For same reaction N2 consumed all.So N2 = 0 at equlibri.

So according to stoichiometry ,ICE table as following:

Reaction O2 N2 2NO Initial mole 0.01 0.01 0 Change ? ? ? Equlibrium ? ? 0.0002

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