What is the rate of the reaction when the initial concentration of no is 0.400 m and that of br2 is 0.265 m ?

 This problem is actually quite simple. The reason it seems so hard is that a lot of useless, extraneous information is given to throw you off the correct track (unless there is a continuation of the problem that you did not give). The question wants the rate of reaction; therefore ignore all information pertaining to equilibrium; there is no relationship between the two. Moreover, the mechanism for the reverse of a reaction that comes to dynamic equilibrium can be completely different for the forward reaction, and so could its rate law, so ignore the given information on the rate constant of the reverse reaction. 

We first must find the rate law of the forward reaction. Comparing trial 2 to trial 1, we see that the initial rate quadruples (4x) when the concentration of NO is doubled. Therefore, the reaction is second order in [NO]. Comparing trial 3 to trial 2, we see that the initial rate is doubled when the concentration of Br2 is doubled, so the reaction is first order in [Br2]. The rate law is: 
Rate = k [NO]² [Br2]. 
In this instance, the rate law follows the stoichiometric coefficients; however, this is not always the case. 

To find the rate constant k of the forward reaction, just substitute the given information from any of the trials into the rate law, and solve for k. Using the first trial: 
29 m/s = k (0.15 M)² (0.240 M); 
solving, k = 5370.4 /M²·s. 

Now use the calculated rate constant and the rate law to find the desired  
Rate = (5370.4 /M²·s) (0.400 M)² (0.265 M) = 227.7 M/s.

Anions are negative charged ions