Calculate the rate constant k.

Answer: 0.20 M⁻¹·s⁻¹ k = rate/[A]² = (2.0×10⁻³)/(0.10)² = 2.0×10⁻³/0.010 = 0.20 M⁻¹·s⁻¹.

What is the rate law?

Answer: rate = k[S2O8^2-][I-] Doubling [S2O8^2-] halves t (doubles rate) - first order. Doubling [I-] also halves t - first order. Overall rate = k[S2O8^2-][I-].

AP Chemistry · Topic 5.2

Introduction to Rate Law Practice

Part of Kinetics.(TRA-3.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$Using the same data set from cluster c4-q1, calculate k.$
$Calculate the rate constant k.$
- A
  $0.20 M⁻¹\cdots⁻¹$
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- B
  $2.0 M⁻¹\cdots⁻¹$
- C
  $0.020 s⁻¹$
- D
  $2.0\times10⁻² M⁻¹\cdots⁻¹$
Why
$k = rate/[A]² = (2.0\times10⁻³)/(0.10)² = 2.0\times10⁻³/0.010 = 0.20 M⁻¹\cdots⁻¹.$
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Sample 2difficulty 2/5
$A rate vs [A] plot is curved upward (rate \propto [A]^n). Doubling [A] makes rate increase 4\times. The order is:$
- A
  $1$
- B
  $2$
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- C
  $0.5$
- D
  $3$
Why
$2^n = 4 -> n = 2.$
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Sample 3difficulty 2/5
$Students study the reaction S2O8^2- + 2 I- -> 2 SO4^2- + I2 using a clock method (starch indicator + thiosulfate). They measure time t for the blue color to appear in three trials at 25 C with constant ionic strength. Trial 1: [S2O8^2-]=0.040 M, [I-]=0.040 M, t=84 s Trial 2: [S2O8^2-]=0.080 M, [I-]=0.040 M, t=42 s Trial 3: [S2O8^2-]=0.040 M, [I-]=0.080 M, t=42 s$
$What is the rate law?$
- A
  $rate = k[S2O8^2-][I-]$
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- B
  $rate = k[S2O8^2-][I-]^2$
- C
  $rate = k[S2O8^2-]^2[I-]$
- D
  $rate = k[S2O8^2-]$
Why
$Doubling [S2O8^2-] halves t (doubles rate) -> first order. Doubling [I-] also halves t -> first order. Overall rate = k[S2O8^2-][I-].$
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Sample 4difficulty 2/5
$A student studies the bleaching of crystal violet (CV+) by NaOH using a spectrophotometer at 590 nm. With NaOH in large excess (pseudo-first-order), the absorbance is recorded versus time: t (s): 0 30 60 90 120 150 A: 0.800 0.610 0.464 0.353 0.269 0.205 A plot of ln(A) versus t is linear with slope = -0.00908 s^-1.$
$Repeating the experiment with twice the [NaOH] gives k_obs = 0.0182 s^-1. What is the order with respect to OH-?$
- A
  $Second order$
- B
  $First order$
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- C
  $One-half order$
- D
  $Zero order$
Why
$Doubling [OH-] doubled k_obs, indicating k_obs proportional to [OH-]^1, so first order in OH-.$
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Sample 5difficulty 2/5
$Overall order of rate = k[A][B]² is$
- A
  $4$
- B
  $3$
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- C
  $1$
- D
  $2$
Why
$Sum of exponents: 1 + 2 = 3.$
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