chemistry review ### Second-Order Kinetics ExampleAt a certain temperature, this reaction follows second-order kinetics with a rate constant of \( 0.0159 \, M^{-1} \cdot s^{-1} \):\[ \text{H}_2\text{CO}_3 (aq) \rightarrow \text{H}_2\text{O} (aq) + \text{CO}_2 (aq) \]Suppose a vessel contains \(\text{H}_2\text{CO}_3\) at a concentration of \(1.04 \, M\). Calculate the concentration of \(\text{H}_2\text{CO}_3\) in the vessel 610 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.### Given Data1. Initial concentration of \(\text{H}_2\text{CO}_3\) (\( [\text{H}_2\text{CO}_3]_0 \)) = 1.04 M2. Rate constant (k) = \( 0.0159 \, M^{-1} \cdot s^{-1} \)3. Time (t) = 610 seconds### Second-Order Kinetics FormulaFor a second-order reaction, the concentration at time \( t \) can be calculated using the formula:\[ \frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0} \]Where:- \([A]_t\) is the concentration at time \( t \)- \( k \) is the rate constant- \( [A]_0 \) is the initial concentration- \( t \) is the time### Calculation1. Substitute the given values into the second-order kinetics formula:\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + \frac{1}{1.04} \]2. Perform the calculation:\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 0.0159 \times 610 + 0.9615 \]\[ \frac{1}{[\text{H}_2\text{CO}_3]_t} = 9.

H2CO3(aq) ----> H2O(aq) + CO2(aq)Reaction follows second order kinetics.Rate constant (k) =…

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.0159M¹¹; H,CO, (aq) →H,O(aq)+CO,(aq) Suppose a vessel contains H₂CO3 at a concentration of 1.04M. Calculate the concentration of H₂CO3 in the vessel 610. seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. M