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Part C:
The reactant concentration in a first-order reaction was 8.80×10−2 M after 50.0 s and 2.70×10−3 M after 100 s . What is the rate constant for this reaction?
Answer: K1st
Part D:
The reactant concentration in a second-order reaction was 0.810 M after 125 s and 5.20×10−2 M after 710 s . What is the rate constant for this reaction?
Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter as N*m.
Answer: K2nd=
Using Integrated Rate Laws?
2016-02-26 12:19 pm
回答 (1)
2016-02-26 12:59 pm
Part C :
Method 1 :
For a first-order reaction : ln[A] = -kt + ln[A]₀
After 50.0 s, [A] = 8.80 × 10⁻² M :
ln(8.80 × 10⁻²) = -k(50.0) + ln[A]₀ ...... [1]
After 100 s, [A] = 2.70 × 10⁻³ M :
ln(2.70 × 10⁻³) = -k(100) + [A]₀ ...... [2]
[1] - [2] :
[ln(8.80 × 10⁻²) - ln(2.70 × 10⁻³)] = 50k
Rate constant, k = 6.97 × 10⁻² s⁻¹
Method 2 :
[A]₀ = 8.80 × 10⁻² M
[A] = 2.70 × 10⁻³ M
t = 100 - 50.0 = 50 s
For a first-order reaction : ln[A] = -kt + ln[A]₀
ln(2.70 × 10⁻³) = -k(50) + ln(8.80 × 10⁻²)
Rate constant, k = 6.97 × 10⁻² s⁻¹
====
Part D:
Part C :
Method 1 :
For a second-order reaction : (1/[A]) = kt + (1/[A]₀)
After 125 s, [A] = 0.810 M :
(1/0.810) = k(125) + (1/[A]₀) ...... [1]
After 710 s, [A] = 5.2 × 10⁻² M = 0.0520 M:
(1/0.0520) = k(710) + (1/[A]₀) ...... [2]
[2] - [1] :
(1/0.0520) - (1/0.810) = 585k
Rate constant, k = 3.08 × 10⁻² M⁻¹*s⁻¹
Method 2 :
[A]₀ = 0.810 M
[A] = 5.20 × 10⁻² M = 0.0520 M
t = 710 - 125 = 585 s
For a second-order reaction : (1/[A]) = kt + (1/[A]₀)
(1/0.0520) = k(585) + (1/0.810)
Rate constant, k = 3.08 × 10⁻² M⁻¹*s⁻¹
Method 1 :
For a first-order reaction : ln[A] = -kt + ln[A]₀
After 50.0 s, [A] = 8.80 × 10⁻² M :
ln(8.80 × 10⁻²) = -k(50.0) + ln[A]₀ ...... [1]
After 100 s, [A] = 2.70 × 10⁻³ M :
ln(2.70 × 10⁻³) = -k(100) + [A]₀ ...... [2]
[1] - [2] :
[ln(8.80 × 10⁻²) - ln(2.70 × 10⁻³)] = 50k
Rate constant, k = 6.97 × 10⁻² s⁻¹
Method 2 :
[A]₀ = 8.80 × 10⁻² M
[A] = 2.70 × 10⁻³ M
t = 100 - 50.0 = 50 s
For a first-order reaction : ln[A] = -kt + ln[A]₀
ln(2.70 × 10⁻³) = -k(50) + ln(8.80 × 10⁻²)
Rate constant, k = 6.97 × 10⁻² s⁻¹
====
Part D:
Part C :
Method 1 :
For a second-order reaction : (1/[A]) = kt + (1/[A]₀)
After 125 s, [A] = 0.810 M :
(1/0.810) = k(125) + (1/[A]₀) ...... [1]
After 710 s, [A] = 5.2 × 10⁻² M = 0.0520 M:
(1/0.0520) = k(710) + (1/[A]₀) ...... [2]
[2] - [1] :
(1/0.0520) - (1/0.810) = 585k
Rate constant, k = 3.08 × 10⁻² M⁻¹*s⁻¹
Method 2 :
[A]₀ = 0.810 M
[A] = 5.20 × 10⁻² M = 0.0520 M
t = 710 - 125 = 585 s
For a second-order reaction : (1/[A]) = kt + (1/[A]₀)
(1/0.0520) = k(585) + (1/0.810)
Rate constant, k = 3.08 × 10⁻² M⁻¹*s⁻¹
收錄日期: 2021-04-18 14:35:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160226041917AA3v8Er