How to do 5a and 5b?

4.
i.
Let a and b be the order of reaction respect to X and Y respectively.
Then, Rate = k [X]ᵃ [Y]ᵇ

Experiment 1: 2.20 × 10⁻⁴ = k (1.0)ᵃ (1.0)ᵇ …… {1}
Experiment 2: 1.70 × 10⁻³ = k (2.0)ᵃ (2.0)ᵇ …… {2}
Experiment 3: 6.60 × 10⁻⁴ = k (1.0)ᵃ (3.0)ᵇ …… {3}
Experiment 4: 3.96 × 10⁻³ = k (3.0)ᵃ (2.0)ᵇ …… {4}

{3} / {1}:
(6.60 × 10⁻⁴) / (2.20 × 10⁻⁴) = [k (1.0)ᵃ (3.0)ᵇ] / [k (1.0)ᵃ (1.0)ᵇ]
3 = 3ᵇ
b = 1

{4} / {2}:
(3.96 × 10⁻³) / (1.70 × 10⁻³) = [k (3.0)ᵃ (2.0)ᵇ] / [k (2.0)ᵃ (2.0)ᵇ]
2.33 = (3/2)ᵃ
1.5ᵃ ≈ 1.5²
a = 2

The order of reaction is 2 with respect to X, and 1 with respect to Y.


ii.
Substitute a = 2 and b = 1 into {1} :
2.20 × 10⁻⁴ = k (1.0)² (1.0)¹
The rate constant, k = 2.20 × 10⁻⁴ M⁻² s⁻¹


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5.
a.
Rate constant, k = ln(2) / (half-life) = ln(2) / (6.58 × 10³ years) = 1.05 × 10⁻⁴ year⁻¹

b)
Number of half-lives for 1000 years = (1000 years) / (6.58 × 10³ years/half-life) = 0.152 half-life
Fraction of sample remain = (1/2)⁰·¹⁵² = 0.9 = 9/10

Radioactive things decay into non-radioactive things. So after a certain time, there is less of the radioactive thing present (because some of it has decayed into something else). The amount left at any time (t) usually follows a certain equation:

t = amount of time passed. If the half life has units of years, then this should also be in years. It has the same units as the half life
A0 = amount of radioactive substance initially present (at time when t = 0)
A(t) = amount of radioactive substance left after time t has passed
h = half life

A(t) = (A0) * ((1/2)^(t/h))

Part b gives you the time passed (t), and h. "Fraction of the sample" means what fraction of the substance is left, which is A(t)/A0. So you don't need to know what A(t) or A0 are. But you are solving for A(t)/A0, (which is the fraction of how much is left)

It seems ambiguous what "rate constant" means. This should be explained in your class. Maybe it is
(1/2)^(1/h)??????????, which is part of the formula: A(t) = (A0) * ((1/2)^(t/h))

which can be written as:

A(t) = (A0) * ((1/2)^(1/h))^t

So if r = (1/2)^(1/h), then
A(t) = (A0) * (r)^t

This explains radioactive decay better:
https://www.khanacademy.org/science/in-in-class-12th-physics-india/nuclei/in-in-half-life-and-decay-rate/v/half-life