Question

6. The reaction NO2(g) + NOR) <===> N200) + O2(g) has Keq = 0.914 at a certain temperature. At the same temperature, what is the value of Keq for the following reaction? 2 N2O(g) + 2 Ozta) <===> 2 NO26) + 2NO() 7. The reaction NO2(g) + NO(g) <===> N2010) + O2(g) has AH = -42.9kJ. a) List the kinds of changes that would alter the amount of NO2 present in the system at equilibrium b) What could be done to this system without changing the amount of NO, at equilibrium? 8. At 100°C, Keq = 0.135 for the reaction: 3 H269) + N2() <===> 2 NH3(e) In a reaction mixture at equilibrium at this temperature, [NH3] = 0.030 M and [N2) = 0,50 M. What is the molar concentration of H2(0)? 9. The reaction N2O3(e) <===> NO(g) + NO269) is to be studied at a certain temperature. A mixture was prepared in which the initial concentrations of NO and NO, were 0.40 and 0.60 M, respectively. When equilibrium was reached, the concentration of N20, was measured to be 0.13 M. What were the equilibrium concentrations of NO and NO2?

i need help with question number 6, 7 a, b, 8, and 9

Answer 1

6. Keq of NO2(g) + NO(g) -------> N2O(g)+ O2(g) is 0.914

keq = [N2O][O2]/[NO2][NO]

so if we multiply th equation by 2 keq will also be raised to power 2 that will be .914^2 = 0.835

7. Keq for equation is keq = [N2O][O2]/[NO2][NO]

As NO2 is in the reactant side, so increaing concentration of any component of product will push the equilibrium backward hence NO2 will increase and vice versa is also true i.e. decreasing concentration of product will push the equilibrium forward hence NO2 will decrease.

Also if we increase the concentration of NO equilibrium will move forward hence NO2 will decrease.

The position of equilibrium also changes if you change the temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favoring the reaction which absorbs heat. So our reaction is exothermic i.e. heat is in the product side so increase in temperature will favour the forward reaction i.e. NO2 will decrease and vice versa.

8. N2​+3H2→2NH3​

Keq = [NH3]2/[N2][H2]3

given Keq = 0.135 [NH3]=0.030 M/l [N2]=0.50 M/l

0.135 = (0.030^2)/(0.5)[H2]3

[H2]3 = (0.030^2)/(0.5*0.135)

[H2] = 0.237 M/L

8. N2O3(g) <=====> NO(g) + NO2(g)

intial conc 0 0.4 0.6

final conc x 0.4-x 0.6-x

for the following question final concentration of N2O3 is given 0.13 so final concentrations of NO will be 0.4-0.13 = 0.27, while that of NO2 is 0.6-0.13=0.47.