Please answer question #3. Thank you! 0.0160 mol/(L s)Determine a set of possible coefficients to balance this general reaction.2. The rate of the reaction between hemoglobin (Hb) and carbon monoxide (CO) was studied at20°C. The following data were collected with all concentration units in umol/L. (A hemoglobinconcentration of 2.21umol/L is equal to 2.21 x 10-6 mol/L.)[Hb], (umol/L )[CO),(µmol/L )Initial Rate (umol/(L s) )2.211.000.6194.421.001.244.423.003.71a. Determine the rate law.b. Calculate the value of the rate constant.c. What would be the initial rate for an experiment with [Hb], = 3.50 µmol/L and [CO], = 2.00umol/L?3. The following initial rate data was collected at 25°C for the reaction:4A+В+С, —>А,B + 2 AСIdentify the rate law for this reaction, the rate constant, and the order of the reaction.expInitial Rate[A][B][C](M/s)#16.8 x 1060.20.30.2#22.7 x 10-50.20.60.2#31.0 x 10-50.30.30.2# 42.7 x 10-50.20.60.64. The reaction M+N –→ 0+Phas a rate law: rate = 3.5 x 10-3 M-'s' [N]?. If the initialconcentration of N is 0.8 M, how many minutes pass before 65% of N has reacted? What is thehalf life (in sec) if the initial concentrations of M and N are 1.3 M and 0.6 M, respectively.5. A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constantand the half-life for this process?

The reaction taking place is given as, => 4 A + B + C2 ---------> A2B + 2 AC NOTE : In the experiment data, last column will be [C2 ] i.e concentration of reactant C2 and not C. Since the rate law of any reaction depends only on the reactant concentrations. Hence the rate law expression for the above reaction can be written as, => Rate = K [A]a [B]b [C2 ]c where K = Rate constant [A] = concentration of A a = order of reaction with respect to A [B] = concentration of B b = order of reaction with respect to B [C2 ] = concentration of C2 c = order of reaction with respect to C2 Substituting the given values into the equation we get, 1) 6.8 X 10-6 = K X (0.2)a X (0.3)b X (0.2)c 2) 2.7 X 10-5 = K X (0.2)a X (0.6)b X (0.2)c 3) 1.0 X 10-5 = K X (0.3)a X (0.3)b X (0.2)c 4) 2.7 X 10-5 = K X (0.2)a X (0.6)b X (0.6)c Dividing equation 2 from equation 1 we get, => 3.97 = 2b => b = 2 approx. Dividing equation 4 from 2 we get, => 1 = 3c => c = 0 Dividing equation 3 from 1 and substituting the value of c we get, => 1.47 = 1.5a => a = 1 approx. Hence the rate of the reaction is => Rate = K [A] [B]2 [C2 ]0 = K [A] [B]2.Substituting the values of a, b and c in equation 1 we get, => 6.8 X 10-6 = K X (0.2)1 X (0.3)2 X (0.2)0 => K = Rate constant = 3.78 X 10-4 M-2s-1 approx. The overall order of the reaction = a + b + c = 1 + 2 + 0 = 3