08 Question (4 points) a See page 648 Nitrogen dioxide reacts with SO2 to form N

Question

08 Question (4 points) a See page 648 Nitrogen dioxide reacts with SO2 to form NO and SO NO2(8)+SO,(g)NO(g)+So(8) An equilibrium mixture is analyzed at a certain temperature and found to contain [NO2]-0.100 M SO2l-0.300 M, [NO]- 2.00 M, and [SO3)-0.600 M. At the same temperature, extra SO2(g) is added to make a total [SO2l-0.800 M 1st attempt see Periodic Table Part 1 (1 point) Calculate the composition of the mixture when equilibrium has been reestablished. [NO2] Part 2 (1 point) Part 3 (1 point) NO] Part 4 (1 point) See Hint [S031-

Explanation / Answer

(8)

The reaction taking place is:

NO2 + SO2 ---> NO + SO3

Calculating the equilibrium conc value:

K = ([NO]*[SO3])/([NO2]*[SO2]) = (2*0.6)/(0.1*0.3) = 40

When more of SO2 is added, reaction will proceed towards the right to adjust itself. Making the ICE table:

NO2 + SO2 ---> NO + SO3

Initial 0.1 0.8 2 0.6

Change -x -x +x +x

Eqb 0.1-x 0.8-x 2+x 0.6+x

Calculating eqb constant:

K = ( (2+x)*(0.6+x) )/( (0.1-x)*(0.8-x) ) = 40

Solving we get:

x = 0.054

So,

[NO2] = 0.1-x = 0.1-0.054 = 0.046 M

[SO2] = 0.8-x = 0.8-0.054 = 0.746 M

[NO] = 2+x = 2+0.054 = 2.054 M

[SO3] = 0.6+x = 0.6+0.054 = 0.654 M

Hope this helps !

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