Question

(10pts) The equilibrium constant, K, is 2.4 x 10' at a certain temperature for the reaction : 2NO2艹Nao + O20 6. For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a) A 1.0-Lflask contains 0.024 mol NO, 2.0 mol Na, and 2.6 mol O b) A 2.0-L flask contains 0.032 mol NO, 0.62 mol N, and 4.0 mol O c) A 3.0-L flask contains 0.060 mol NO, 2.4 mol Ne, and 1.7 mol O

Answer 1

K = N2*O2 / (NO2)^2

in order to be in equilbirium

Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.

For this, we use Q, the reaction quotient of products/reactants, it allows us to understand the ratio distribution and the direction/shit of equilibrium

Q is defined as:

Q = [C]^c * [D]^d / ([A]^a * [B]^b)

In this Case, the concentrations are NOT in equilibrium

Therefore:

If Q < Keq; this has much more reactants than products, therefore expect reactants to form more product in order to achieve equilibrium

If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium

If Q = Keq; this has the same ratio in equilibrium for reactants and products. Expect no reaction. It is safe to assume this is already in equilibrium.

then

a)

Q = [N2][O2]/[NO]^2

Q= (2*2.6)/(0.024^2) = 9027; not K

shift goes to reactants

b)

Q = [N2][O2]/[NO]^2

Q= (0.62*4)/(0.032^2) = 2421.875;

pretty near to K, ths must ne in equilbirium

c)

Q = [N2][O2]/[NO]^2

Q= (2.4*1.7)/(0.06^2) = 1133.33 ;

not Q =/ K

therefore, not in equilbirium

shift goes to productrs