In a first-order decomposition reaction. 58.6% of a compound decomposes in 11.7 min. How long (in min) does it take for 80.2% of
1 answer:
Answer:
t = 21.5 min.
Explanation:
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In this case, since the kinetics of a first-order reaction is:
Thus, since we are given the 11.7 min for a 58.6-% consumption, we can compute the rate constant, k:
Now, for the second problem, as the new consumption is 80.2%, we can compute the required time as shown below:
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Explanation:
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Answer:
The answer to the question is
The rate constant for the reaction is 1.056×10⁻³ M/s
Explanation:
To solve the question, e note that
For a zero order reaction, the rate law is given by
[A] = -k×t + [A]₀
This can be represented by the linear equation y = mx + c
Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀
Therefore the rate constant k which is the gradient is given by
Gradient = where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M
= = -0.001056 M/s = -1.056×10⁻³ M/s
Threfore k = 1.056×10⁻³ M/s