2. The reaction is expressed by the equation: A+B=C At initial concentratio ns of C_(0)(A)=3mol// and C_(0)(B)=5mol/l the reaction rate is 0.3mol/(lcdot s) Determine a) the rate constant b) the reaction rate after time when the concentration of A decreases by 2mol/I

To determine the rate constant and the reaction rate after a certain time, we need to use the given information and the rate law for the reaction.Given information:- Reaction equation: A + B = C- Initial concentrations: and - Reaction rate: a) To determine the rate constant, we can use the rate law for the reaction. The rate law for a reaction is given by:Rate = k[A][B]where k is the rate constant, [A] and [B] are the concentrations of the reactants A and B, respectively.We can rearrange the rate law to solve for the rate constant k:k = Rate / ([A][B])Substituting the given values, we have:k = 0.3 mol/(L·s) / ((3 mol/L)(5 mol/L))k = 0.02 L^2/mol·sTherefore, the rate constant is 0.02 L^2/mol·s.b) To determine the reaction rate after a certain time when the concentration of A decreases by 2 mol/L, we need to use the integrated rate law for the reaction.The integrated rate law for a first-order reaction is given by:ln([A]/[A0]) = -ktwhere [A0] is the initial concentration of A, [A] is the concentration of A at time t, and k is the rate constant.Given that the concentration of A decreases by 2 mol/L, we can calculate the new concentration of A:[A] = [A0] - 2 mol/L[A] = 3 mol/L - 2 mol/L[A] = 1 mol/LSubstituting the values into the integrated rate law, we have:ln(1 mol/L / 3 mol/L) = -0.02 L^2/mol·s * tSolving for t, we get:t ≈ 50 sTherefore, the reaction rate after 50 seconds when the concentration of A decreases by 2 mol/L is 0.3 mol/(L·s).