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# Even%20More%20Rate%20Law%20Answers - [0.010 2[1.50 2 = 3.33...

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Even More Rate Law Practice 1. Given the following experimental data, develop a rate law: Rate (mol/L-s) A (mol/L) B (mol/L) C (mol/L) 8.4342 * 10 -10 0.0718 0.0585 0.129 9.3714 * 10 -11 0.0718 0.0585 0.043 1.3495 * 10 -8 0.6462 0.234 0.129 8.4342 * 10 -10 0.6462 0.0585 0.129 Rate = k [B] 2 [C] 2 k = 1.481 * 10 −5 2. Solve for the units of the rate constant k: Fourth order overall, k = L 3 / (mol 3 s) 3. Solve for the rate with the following concentrations [A] = 0.015, [B] = 0.010, [C] = 1.50 Rate = 1.481 * 10 −5
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Unformatted text preview: [0.010] 2 [1.50] 2 = 3.33 * 10 −9 mol / (L s) 4. Given that the chemical reaction is: 2 A + 2 B + 3 C 4 D, solve for the rate of A in the conditions described in question 3. 3.33 * 10 −9 mol / (L s) = (1/2) Rate(A) Rate(A) = 6.66 * 10 −9 mol / (L s) 5. Using the above information, solve for the rate of D, in the conditions described in question 3. 3.33 * 10 −9 mol / (L s) = (1/4) Rate(D) Rate(D) = 1.33 * 10 −8 mol / (L s)...
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