Question

The generic reaction has the following rate laws: 2A=B forward reaction : rate = kf[A]^2 reverse reaction : rate = kr[B] where kf is the rate constant for the forward reaction and k1 is the rate constant for the reverse reaction. At equilibrium, The two rates are equal and so kf[A]^2 = kr[B]. The equilibrium constant for a reaction is related by the law of mass action to the rate constants for the forward and reverse reactions: Formation of nitrosyl bromide Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with bromine. Br2The reaction rapidly establishes equilibrium when the reactants are mixed. Part A At a certain temperature the concentration of NO was 0400 M and that of Br2 was 0.23SM At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature? Express your answer numerically.Kc=0.1058 Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s).

I really wish I knew where to start on this one, any help would be awesome!

Answer 1

Answer:

2NO (g) + Br2 (g) <====> 2NOBr

Initial concentration of NO = 0.400 M
Initial concentration of Br2 = 0.235 M
Initial concentration of NOBr = 0 M

We know that the final concentration of NOBr = 0.250 M

Therefore,
Change in concentration of NOBr = + 0.250 M
Change in concentration of NO = - 0.250 M (Since no. of mole of NO = no. of mole of NOBr)
Change in concentration of Br2 = - 0.125 M (Since no. of mole of Br2 = 0.5 * no. of mole of NOBr)

Final concentration of NO = 0.400 - 0.250 = 0.150 M
Final concentration of Br2 = 0.235 - 0.125 = 0.110 M

Kc = conc.(NOBr)^2 / [ (conc. (NO)^2) * (conc. (Br2) ]
=(0.250^2)/( 0.150^2 * 0.110)
=15.15

Kc= 15.15 mol