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Rates of Reactions

Rates of Reactions. Learning Goals. Students will: understand the Rate Law Equation determine the Rate Law Equation given experimental data. Success Criteria. Students will: Be able to write Rate Law expressions for chemical reactions. Purpose. We understand:

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Rates of Reactions

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  1. Rates of Reactions

  2. Learning Goals Students will: • understand the Rate Law Equation • determine the Rate Law Equation given experimental data

  3. Success Criteria • Students will: • Be able to write Rate Law expressions for chemical reactions

  4. Purpose We understand: • How the rate of a reaction changes during a reaction (the rate decreases as reactants are used up) • How to calculate the rate of reaction based on data from a graph. (r = ∆c/∆t) • which factors affect the rates of a reaction. What we want to do: • Is develop a mathematical equation for rates of reaction (r) that incorporates some rate factors.

  5. The Rate Law or Law of Mass Action • The rate of a chemical reaction is proportional to the product of the concentrations of the reactants • OR The rate, r, will always be proportional to the product of the initial concentrations of the reactants, where these concentrations are raised to some exponential values. • For a typical chemical reaction: a A + b B (products) This can be expressed as r α [A]m[B]n Where the exponents m and n are “orders of reaction” These exponentsare NOT related to the co-efficients

  6. Rate Law Equation • The relationship r α [A]m[B]n can be converted to the Rate Law equation: r = k [A]m[B]n Where: • [A] and [B] represent the concentrations of substance A and substance B • k = rate law constant (which is not affected by the concentration) • m and n are the “orders of reaction”

  7. Determining Orders of Reaction • The values m and n MUST be determined empirically (by conducting an experiment and mathematically analyzing the data) • The exponents, m and n may be zero, fractions or integers. • Some “experimental” data is provided below for the following equation: 2 X + 2 Y + 3 Z → products • This means the rate law equation will be: r = k[X]m[Y]n[Z]p • Let’s determine m, n and p.

  8. 2 X+ 2 Y + 3 Z → products • Determine how the rate of reaction changes due to changes in concentration of X ([X]). • To do this, we must keep the concentrations of Y and Z constant – they are controlled variables. Any changes in r must be entirely due to X. Experimental data showed that: • As the concentration of X is doubled (x2), the rate of reaction also doubles (x2) • As the concentration of X is tripled (x3), the rate of reaction also triples (x3)

  9. Determining m • We have discovered a linear relationship. • r  [X]1 (y = x1) • Linear relationships are known as first order relationships because the exponent is 1. • Since the exponent is 1, then m=1

  10. 2 X + 2 Y + 3 Z → products • Determine how the rate of reaction changes due to changes in concentration of Y ([Y]). • To do this, we must keep the concentrations of X and Z constant – they are controlled variables. Any changes in r must be entirely due to Y. Experimental data showed that: • As the concentration of Y is doubled (x2), the rate of reaction multiplies by 4 (x4) (4 = 22) • As the concentration of Y is tripled (x3), the rate of reaction multiplies by 9 (x9) (9 = 32)

  11. Determining n • We have discovered an exponential relationship. • r  [Y]2 (y = x2) • Linear relationships are known as second order relationships because the exponent is 2. • Since the exponent is 2, then n=2

  12. 2 X + 2 Y+ 3 Z → products • Determine how the rate of reaction changes due to changes in concentration of Z([Z]). • To do this, we must keep the concentrations of X and Y constant – they are controlled variables. Any changes in r must be entirely due to Z. Experimental data showed that: • As the concentration of Z is doubled (x2), the rate of reaction does not change (x1) (1 = 20) • As the concentration of Z is tripled (x3), the rate of reaction does not change (x1) (1 = 30)

  13. Determining p • We have discovered an line relationship. • r  [Z]0 (y = xo) • Line relationships are known as zeroth order relationships because the exponent is 0. • Since the exponent is 0, then p=0

  14. Rewrite the Rate Law • r = k[X]1[Y]2[Z]0 • The overall order of reaction is determined by adding the individual orders of reaction. • In this case 1 + 2 + 0 = 3

  15. Let’s use some real data! • Start by writing the Rate Law equation • r = k[BrO3-(aq)]m[HSO3-(aq)]n • Determine how the rate of reaction changes due to changes in concentration of BrO3- ([BrO3-(aq)]). • To do this, we must keep the concentration of HSO3- constant ([HSO3-(aq) ]) – it is a controlled variable. Any changes in r must be entirely due to BrO3-(aq). • Let’s look at the data again – are there any 2 trials we can use for comparison in which [BrO3-(aq)] changes and [HSO3-] remains constant.

  16. Determine m and n • Let’s choose trials 2 and 3. [BrO3-(aq)] changes and [HSO3-] remains constant. • As the concentration of [BrO3-(aq)]is doubled (x2), the rate of reaction also doubles (x2) (2 = 21) • This is a first-order reaction, m = 1

  17. Determine m and n • Determine how the rate of reaction changes due to changes in concentration of HSO3-([HSO3-(aq) ]). • To do this, we must keep the concentration of BrO3-constant ([BrO3-(aq)]) – it is a controlled variable. Any changes in r must be entirely due to HSO3-(aq). • Let’s look at the data again – are there any 2 trials we can use for comparison in which [HSO3-]changes and [BrO3-(aq)]remains constant.

  18. Determine m and n • Let’s choose trials 1 and 2.[HSO3-] changes and [BrO3-(aq)] remains constant. • As the concentration of [HSO3-]is doubled (x2), the rate of reaction multiplies by 4 (x4) (4 = 22) • This is a second-order reaction, m = 2 • Rewrite the Rate Law Equation; r = k[BrO3-(aq)]1[HSO3-(aq)]2

  19. Now let’s determine k(the Rate Law constant) • Since r = k[BrO3-(aq)]1[HSO3-(aq)]2 • Then k = r [BrO3-(aq)]1[HSO3-(aq)]2 • Go back to the data and choose any trial. I will use trial 1. Input the r, [BrO3-(aq)], and [HS.O3-(aq)]values. • Therefore: k = 0.20 mmol/L·s [2.0 mmol/L]1[3.0 mmol/L]2 k = 0.011 L2/mmol2·s • Now:r = 0.011 L2/mmol2· s[BrO3-(aq)]1[HSO3-(aq)]2 • See page 376 for tips on the units for k

  20. Applications • Now that we have a completed rate Law Equation: r = 0.011 L2/mmol2· s[BrO3-(aq)]1[HSO3-(aq)]2 • We can input any concentrations of the reactants and determine a rate of reaction. • Try This: What is the rate of reaction if: [BrO3-(aq)] = 0.10 mmol/L and [HSO3-(aq)]= 0.10 mmol/L?

  21. Review of Order of Reaction

  22. Review of Order of Reaction

  23. Determine the Rate Law Equation for these sets of data

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