Let's break down the problem step by step: --- ### **Step 1: Write the reaction and initial conditions** **Reaction:** \[ \text{HCHO} + \text{H}_2\text{O}_2 \rightarrow \text{HCOOH} + \text{H}_2\text{O} \] - **HCHO:** 200 mL, 1 M - **H₂O₂:** 100 mL, 2 M **Total Volume = 200 mL + 100 mL = 300 mL = 0.3 L** --- ### **Step 2: Calculate Initial Moles and Concentrations in Mixture** - **HCHO:** \( 200 \text{ mL} \times 1 \text{ M} = 0.200 \text{ mol} \) - **H₂O₂:** \( 100 \text{ mL} \times 2 \text{ M} = 0.200 \text{ mol} \) **Initial concentrations after mixing:** \[ [\text{HCHO}]_0 = \frac{0.200}{0.3} = 0.667 \text{ M} \] \[ [\text{H}_2\text{O}_2]_0 = \frac{0.200}{0.3} = 0.667 \text{ M} \] Both are equal; this is a **1:1 reaction**. --- ### **Step 3: Use Data Given After 2 Hours** After 2 h, \([\text{HCOOH}]=0.114 \text{ M}\). So, \(x = 0.114\) M of HCHO and H₂O₂ have reacted. Remaining HCHO and H₂O₂: \[ [\text{HCHO}] = [\text{H}_2\text{O}_2] = 0.667 - 0.114 = 0.553 \, \text{M} \] --- ### **Step 4: Determine Kinetics (Assume 1st Order in Each Reactant)** For a reaction: \[ A + B \to \text{products} \] with equal initial concentration \([A] = [B]\), integrated rate law is: \[ \frac{1}{[A]_0 - x} = \frac{1}{[A]_0} + kt \] But more commonly, for equal initial concentrations (\([A]_0 = [B]_0 = a\)): \[ a - x = \text{concentration at time } t \] \[ \frac{a}{a-x} = e^{kt} \] \[ \ln\left(\frac{a}{a-x}\right) = kt \] --- ### **Step 5: Calculate Rate Constant \(k\)** Given: - \(a = 0.667 \text{ M}\) - \(x = 0.114 \text{ M}\) after \(t = 2 \text{ h}\) \[ \ln\left(\frac{0.667}{0.667 - 0.114}\right) = k \times 2 \] \[ \ln\left(\frac{0.667}{0.553}\right) = 2k \] \[ \ln(1.206) = 2k \] \[ 0.187 = 2k \] \[ k = 0.0936 \ \text{h}^{-1} \] --- ### **Step 6: When is the Reaction 99% Complete?** 99% complete means \(x = 99\%\) of initial, so final concentration: \[ x = 0.99 \times 0.667 = 0.660 \text{ M} \] So, \((a-x) = 0.667 - 0.660 = 0.007 \text{ M}\) \[ \ln\left(\frac{0.667}{0.007}\right) = k t \] \[ \frac{0.667}{0.007} = 95.29 \] \[ \ln(95.29) = k t \] \[ 4.556 = k t \] \[ t = \frac{4.556}{0.0936} = 48.7 \text{ h} \] --- ### **Step 7: Final Concentration of HCOOH** At 99% completion, \(x = 0.660 \) M. This is also the concentration of HCOOH. --- ## **Final Answers:** ### **(1) Time for 99% completion:** \[ \boxed{48.7 \ \text{hours}} \] ### **(2) Concentration of HCOOH at that time:** \[ \boxed{0.660 \ \text{M}} \] --- **Summary Table:** | Step | Quantity | Value | |------|------------------------|------------------| | 1 | Initial [HCHO], [H₂O₂] | 0.667 M each | | 2 | [HCOOH] after 2 h | 0.114 M | | 3 | Rate constant, \(k\) | 0.0936 h⁻¹ | | 4 | Time for 99% complete | 48.7 hours | | 5 | [HCOOH] at 99% | 0.660 M |

**Step 1: Determine initial concentrations and reaction details** The initial moles are 0.200 mol HCHO and 0.200 mol H₂O₂ in 0.3 L, giving initial concentrations of approximately 0.667 M each; these react in a 1:1 ratio, forming HCOOH. **Final answer:** Initial concentrations: [HCHO] = 0.667 M, [H₂O₂] = 0.667 M. --- **Step 2: Calculate the rate constant \(k\) using the data after 2 hours** After 2 hours, [HCOOH] = 0.114 M, so 0.114 mol reacted; applying the integrated first-order rate law for equal initial concentrations yields \(k \approx 0.0936\, \text{h}^{-1}\). **Final answer:** \(k \approx 0.0936\, \text{h}^{-1}\). --- **Step 3: Find the time for 99% reaction completion and final HCOOH concentration** At 99% completion, remaining [HCHO] ≈ 0.007 M; solving for time gives approximately 48.7 hours, and the concentration of HCOOH at that point is 0.660 M. **Final answer:** Time ≈ 48.7 hours; [HCOOH] ≈ 0.660 M.