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loris [4]
3 years ago
5

The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process

with a rate constant of 6.5 1/hours. What is the half-life in hours of the breakdown reaction? Enter to 4 decimal places.
Chemistry
1 answer:
denis23 [38]3 years ago
7 0

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

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<h3><u>Answer and explanation</u>;</h3>
  • To balance the charges of ions the number of electrons lost is equal to the number of electrons gained. The overall net charge must be zero.
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FILL IN THE BLANKWORD BANK (can use more than once): less, increases, decreases, greaterNO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)1.Disturbin
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Explanation:

If we have the following reaction at equilibrium:

<em>                                           aA + bB ⇄ cC + dD</em>

where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B,  C, and D. For the reaction at a particular temperature:

                       Kc=([C]^c *[D]^d)/([A]^a *[B]^b)

where Kc is the equilibrium constant, which holds that <em>for a reversible reaction at equilibrium  and a constant temperature, a certain ratio of reactant and product concentrations has  a constant value, Kc</em> (the equilibrium constant). Note that although the concentrations  may vary, as long as the reaction in in equilibrium and temperatura don't change, the value of <u>K remains constant.</u>

For reactions that have not reached equilibrium, we obtain the reaction quotient (Qc), instead of the equilibrium  constant <u>by substituting the initial concentrations into the equilibrium constant expression.</u>

                        Qc=([Co]^c *[Do]^d)/([Ao]^a *[Bo]^b)

To determine the direction in wich the net reaction will proceed to reach equilibrium, que compare the values of Qc and Kc.

  • Qc < Kc: To reach equilibrium, reactants must be converted to products (→)
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  • Qc > Kc: To reach equilibrium, products must be converted to reactants (←)

Solution:

We have the following reaction:

                             NO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)

So:

Kc=([NOCl]^1*[NO2]^1)/([NO2Cl]^1 *[NO]^1)

     =([NOCl][NO2])/([NO2Cl][NO])  

1. In the equation above, [NO2Cl] it's in the denominator, so if we increase it's numericall value by adding NO2Cl <u>decreases Qc  to a value less than Kc.</u>

<em>(From the chemical point of view, if we disturb the equilibrium adding NO2Cl (a reactant), to reach equilibrium again the system proceeds from left to right (→) consuming this reactant.)</em>

2. To reach a new state of equilibrium (<em>where Qc = Kc</em>), Qc therefore  increases wich means that the denominator of the expression for Qc  decreases <em>(in order to increase the denominator as mention above).</em>

3. To accomplish this, the concentration of reageants decreases <em>(reagents are being consumed), </em>and the concentration of prodcuts increases <em>(products are being formed).</em>

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Hope it helps have a blessed day!:)


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