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Give step-by-step solution with explanation and final answer:Auic-phasa reaction 2A. — Pi tuded in a laboratory baker. Th folowing data cio Ato of Aa sh it [Tomet 0 [0 [0 [0 [10 10 [10 210] Cy (mol 1] ocermine he Si Toa RE Ten ec i pO SEE SE I YN ——— Sn: Ein Ras Detain St (RDS) bts 4 3 REAR deriving global rate laws. Derive the relationship between the heat of reaction AH x, and the ache scagoe lh ovr ese smo Oana Reker tion ei i ER AB ot Bi ETE SY. EC A ———— i Tt wd a et on; 40208 cu CE (e-2) Find the equilibrium conversion X.. and the time required to reach 80% of the equilibrium a Mf ah is A LSA 5 Sd i a ian aecivty Sus. Ota th rou perature concanvaton coi to maize he SH hic Fre sd Pat hi erties Fito ctf kh ee et Hl i ideo pac

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Give step-by-step solution with explanation and final answer:Uploaded ImageAuic-phasa reaction 2A. — Pi tuded in a laboratory baker. Th folowing data cio Ato of Aa sh it [Tomet 0 [0 [0 [0 [10 10 [10 210] Cy (mol 1] ocermine he Si Toa RE Ten ec i pO SEE SE I YN ——— Sn: Ein Ras Detain St (RDS) bts 4 3 REAR deriving global rate laws. Derive the relationship between the heat of reaction AH x, and the ache scagoe lh ovr ese smo Oana Reker tion ei i ER AB ot Bi ETE SY. EC A ———— i Tt wd a et on; 40208 cu CE (e-2) Find the equilibrium conversion X.. and the time required to reach 80% of the equilibrium a Mf ah is A LSA 5 Sd i a ian aecivty Sus. Ota th rou perature concanvaton coi to maize he SH hic Fre sd Pat hi erties Fito ctf kh ee et Hl i ideo pac

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Asked by: vemuri swathi
Created at: 2025-12-21 05:20:27

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Let's solve thefirst part** (reaction order and rate constant) step by step: --- ## **Step 1: Identify the Order of Reaction** Given: The reaction is \(2A \rightarrow P\). Data for \(C_A\) and time are given. Let's test for **second-order** (since the stoichiometry suggests it might be second-order): For a second-order reaction: \[ \frac{1}{C_A} - \frac{1}{C_{A}} = kt \] Let's tabulate \(1/C_A\) as given: | Time (s) | \(C_A\) (mol/L) | \(1/C_A\) (L/mol) | |----------|------------------|-------------------| | | 1.00 | 1.00 | | 30 | .80 | 1.25 | | 60 | .67 | 1.49 | | 90 | .60 | 1.67 | | 120 | .50 | 2.00 | | 150 | .43 | 2.33 | | 180 | .36 | 2.78 | | 210 | .30 | 3.33 | --- ## **Step 2: Check for Linearity** For a second-order reaction, a plot of \(1/C_A\) vs. \(t\) should be linear. - The increase in \(1/C_A\) is roughly proportional to \(t\), so **second-order** seems correct. --- ## **Step 3: Determine the Rate Constant \(k\)** The integrated rate law is: \[ \frac{1}{C_A} - \frac{1}{C_{A}} = kt \] Let's use two points to calculate \(k\): - At \(t = \): \(C_{A} = 1.00 \rightarrow 1/C_{A} = 1.00\) - At \(t = 210\) s: \(C_A = .30 \rightarrow 1/C_A = 3.33\) \[ 3.33 - 1.00 = k \cdot 210 \] \[ 2.33 = k \cdot 210 \] \[ k = \frac{2.33}{210} = .0111\ \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \] Let's check with another point for confirmation (at \(t = 120\) s, \(1/C_A = 2.00\)): \[ 2.00 - 1.00 = k \cdot 120 \] \[ 1.00 = k \cdot 120 \] \[ k = \frac{1.00}{120} = .00833\ \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \] The values are close, but not exactly the same due to small experimental errors or rounding in the table. The average of these two: \[ k_{\text{avg}} = \frac{.0111 + .00833}{2} = .0097\ \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \] You may typically use all data points and linear regression for more accuracy. However, **.009–.011 L·mol⁻¹·s⁻¹** is a good estimate. --- ## **Final Answers** ### **Order of Reaction:** **Second Order** ### **Rate Constant (\(k\)):** \[ k \approx .01\ \text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1} \] (Rounded to two significant digits) --- **Summary Table:** | Quantity | Value | |------------------|-------------------------------| | Reaction Order | Second Order | | Rate Constant, k | \(.01\ \text{L mol}^{-1}\text{s}^{-1}\) | --- If you need the step-by-step for the next question, let me know!

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