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.<br /><br />Given information:<br />- Reaction equation: A + B = C<br />- Initial concentrations: $C_{0}(A) = 3 \, \text{mol/L}$ and $C_{0}(B) = 5 \, \text{mol/L}$<br />- Reaction rate: $0.3 \, \text{mol/(L·s)}$<br /><br />a) To determine the rate constant, we can use the rate law for the reaction. The rate law for a reaction is given by:<br /><br />Rate = k[A][B]<br /><br />where k is the rate constant, [A] and [B] are the concentrations of the reactants A and B, respectively.<br /><br />We can rearrange the rate law to solve for the rate constant k:<br /><br />k = Rate / ([A][B])<br /><br />Substituting the given values, we have:<br /><br />k = 0.3 mol/(L·s) / ((3 mol/L)(5 mol/L))<br />k = 0.02 L^2/mol·s<br /><br />Therefore, the rate constant is 0.02 L^2/mol·s.<br /><br />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.<br /><br />The integrated rate law for a first-order reaction is given by:<br /><br />ln([A]/[A0]) = -kt<br /><br />where [A0] is the initial concentration of A, [A] is the concentration of A at time t, and k is the rate constant.<br /><br />Given that the concentration of A decreases by 2 mol/L, we can calculate the new concentration of A:<br /><br />[A] = [A0] - 2 mol/L<br />[A] = 3 mol/L - 2 mol/L<br />[A] = 1 mol/L<br /><br />Substituting the values into the integrated rate law, we have:<br /><br />ln(1 mol/L / 3 mol/L) = -0.02 L^2/mol·s * t<br /><br />Solving for t, we get:<br /><br />t ≈ 50 s<br /><br />Therefore, the reaction rate after 50 seconds when the concentration of A decreases by 2 mol/L is 0.3 mol/(L·s).