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alex41 [277]
3 years ago
9

A certain first-order reaction is 45.0% complete in 65 s. What are the values of the rate constant and the half-life for this pr

ocess
Chemistry
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

0.01228s⁻¹ = rate constant

Half-life = 56.4s

Explanation:

The first order reaction follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%</em>

ln[45%] = -k*65s + ln[100%]

-0.7985 = -k*65s

0.01228s⁻¹ = rate constant

Half-life is:

Half-life = ln2 / k

<h3>Half-life = 56.4s</h3>

<em />

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Answer: The pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

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K_{a} = 1.0 \times 10^{-8}

Let us assume that amount of Cu(H_{2}O)^{2+}_{6} dissociates is x. So, ICE table for dissociation of  Cu(H_{2}O)^{2+}_{6}  is as follows.

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Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of K_{a} is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

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The expression for K_{a} value is as follows.

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pH = -log [H^{+}]

Substitute the values into above formula as follows.

pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49

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