Introduction to Rate Law

AP Chemistry· difficulty 3/5

Students study the reaction S2O8^2- + 2 I- -> 2 SO4^2- + I2 using a clock method (starch indicator + thiosulfate). They measure time t for the blue color to appear in three trials at 25 C with constant ionic strength. The fixed amount of thiosulfate used means the average rate over time t corresponds to consumption of 1.00e-4 mol/L of S2O8^2-. Trial 1: [S2O8^2-]=0.040 M, [I-]=0.040 M, t=84 s Trial 2: [S2O8^2-]=0.080 M, [I-]=0.040 M, t=42 s Trial 3: [S2O8^2-]=0.040 M, [I-]=0.080 M, t=42 s

What is the value of the rate constant k (Trial 1)?

  • A

    3.0e-2 M^-1 s^-1

  • B

    1.0e-4 M^-1 s^-1

  • C

    7.4e-4 M^-1 s^-1

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  • D

    1.2e-3 s^-1

Explanation

rate = 1.00e-4 / 84 = 1.19e-6 M/s. k = rate / ([S2O8^2-][I-]) = 1.19e-6 / (0.040 * 0.040) = 7.4e-4 M^-1 s^-1.

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