Answer:
a) Half life of the decomposition = 4951.1 s ≈ 4950 s
b) Time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900 s
c) If the initial concentration of SO₂Cl₂ is 1.00 M, time it will take for the concentration to decrease to 0.78 M is 1775s
d) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.146 M
e) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.1398 M
Explanation:
Let C₀ represent the initial concentration of SO₂Cl₂
And C be the concentration of SO₂Cl₂ at anytime.
a) Rate of a first order reaction is represented by
dC/dt = - KC
dC/C = - kdt
Integrating the left hand side from C₀ to C₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(C₀/2)/C₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
t₍₁,₂₎ = (In 2)/k
t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)
t₍₁,₂₎ = 4951.1 s
b) dC/C = - kdt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = - kt
C/C₀ = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C = 25% of C₀ = 0.25C₀
0.25C₀ = C₀ e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
- kt = In 0.25
- kt = - 1.386
t = 1.386/(1.4 × 10⁻⁴) = 9900 s
c) C = C₀ e⁻ᵏᵗ
C = 0.78 M; C₀ = 1.00 M
0.78 = 1 e⁻ᵏᵗ
e⁻ᵏᵗ = 0.78
- kt = In 0.78
- kt = - 0.2485
t = 0.2485/(1.4 × 10⁻⁴) = 1775 s
d) C = C₀ e⁻ᵏᵗ
C₀ = 0.150 M, t = 2 × 10² s = 200 s
C = C₀ e⁻ᵏᵗ
e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972
C = 0.15 × 0.972 = 0.146 M
e) C = C₀ e⁻ᵏᵗ
C₀ = 0.150 M, t = 5 × 10² s = 500 s
C = C₀ e⁻ᵏᵗ
e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324
C = 0.15 × 0.9324 = 0.1398 M
