6. At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order

Question

6. At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the rate constant for the reaction at 150°C is 0.029 min-1 , how long does it take a concentration of 0.050 mol L-1 of acetaldehyde to reduce to a concentration of 0.040 mol L-1 ?

7. The decomposition of hydrogen iodide into hydrogen and iodine is a second order reaction. The rate constant k = 0.080 L mol-1 s-1 . How long does it take an initial concentration of 0.050 M to decrease to half this concentration?

8. The gold-198 isotope has a half-life of 2.7 days. If you start with 10 mg at the beginning of the week, how much remains at the end of the week, seven days later?

PLEASE SHOW FULL WORK. Thanks

Explanation / Answer

6) For a first order reaction integrated rate law expression is given as

ln([A]/[A0]) = –kt ………………….(1)

Where,

[A0] = Initial concentration of reactant.

[A] = Final concentration (amount left) after time ‘t’

k = Rate constant or first order rate constant.

Decomposition of acetaldehyde is first order reaction and given that,

[A0] = 0.050 M/L, [A] = 0.040 M/L, k = 0.029 min–1.

And asked for t = ?

Let us substitute the known values in eq. (1) and solve it for t,

ln(0.040/0.050) = – 0.029 x t

– 0.2231 = – 0.029 x t

0.2231 = 0.029 x t

t = 0.2231/0.029

t = 7.7 min ………………(As we used rate constant in the unit minute)

t = 7 min 42 sec (appr.)

It will take 2 min 24 sec to reduce the acetaldehyde concentration from 0.050 M/L to 0.040 M/L.

============================================================

7) For a second order reaction rate constant (k), Initial concentration [A0] and Half-life time is related T½ as,

T½ = 1/k[A0] ……………….(2)

Hydrogen iodide decomposition follows Second order kinetics and given that,

k = 0.080 LM–1s–1, [A0] = 0.050 M

And asked the time required to decompose HI to half concentration i.e. T½ = ?

We can use eq. (2)

T½ = 1/(0.080 x 0.050)

T½ = 250 s

It will 250 s to decompose HI to half of its initial concentration.

===================================================================

8) Radioactive decay follows first order kinetics.

And Integrated rate law expression is given as,

ln([A]/[A0]) = –t

and = 0.693/T½, with this above equation takes form,

ln([A]/[A0]) = –0.693t/T½ ………………….(3)

Where,

[A0] = Initial concentration of radioactive species.

[A] = Final concentration (amount left) after time ‘t’

= Rate constant or first order rate constant.

T½ = Half-life of decay processes.

For radioactive decay of Gold Au–198,

T ½ = 2.7 days, t = 7 days, [A0] = 10 mg, [A] = ?

Using eq. (3)

ln([A]/10) = –0.693 x 7 /2.7

ln([A]/10) = –1.797

[A]/10 = e–1.797

[A] = 10 x 0.166

[A] = 1.66 mg

1.66 mg Gold-198 will be left at the end of 7th day.

=====================XXXXXXXXXXXXXXXXXXXX========================

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.