What is the rate of the reaction when the initial concentration of no is 0.400 m and that of br2 is 0.245 m ?

If the final concentration of NOBr = 0.250M
Change in concentration of NOBr = +0.250 M
Change in concentration of NO = -0.250 M
Change is concentration of Br2 = -0.125 M ( since number of moles of Br2 is 0.5 times the number of mole of NOBr)
Final concentration of NO = 0.400-0.250 = 0.150 M
Final concentration of Br2 = 0.245 - 0.125 M = 0.0120 M
Therefore;
Kc = conc (NOBr)^2/ ((conc NO)^2 ×(conc (Br2)
     = (0.250²)/(0.150²×0.120)
      = 23.148
      = 23.1 mol/dm³

Ibelieve it is b. proton.hope my answer has you : )