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3 years ago

9

For some transformation having kinetics that obey the Avrami equation (Equation 11.17), the parameter n is known to have a value

of 2.3. If, after 145 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 88% completion

1 answer:

Alex787 [66]3 years ago

3 0

Answer:

t ≈ 235.47 secs

Explanation:

<u>Calculate how long it will take the transformation to go to 88% completion</u>

Given that Avrami equation ( <em>y </em>)= 1 - exp( - kt^n )

n = 2.3

t = 145 secs

reaction = 50% after t = 145 secs

Reaction = 88% after t = ?

attached below is a detailed solution

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Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

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Will be written as:

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}$

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(b) $Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)$

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$Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)$

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(d) $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

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1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o

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Answer:

* $pH=11.0$

* $[HClO]_{eq}=9.32x10^{-4}M$

* $[HF]_{eq}=5.07x10^{-6}M$

Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

$M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M$

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

$NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-$

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

$ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\$

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

$Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}$

In such a way, the law of mass action for each case is:

$Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}$

Now, by solving for the change $x$ for both the ClO⁻ and F⁻ equilibriums, we obtain:

$x_{ClO}=9.32x10^{-4}M\\x_{F}=5.07x10^{-6}M$

Such results equal the concentrations of OH⁻ in each equilibrium, thus, the total concentration of OH⁻ result:

$[OH^-]_{tot}=9.32x10^{-4}M+5.07x10^{-6}M=9.37x10^{-4}M$

With which the pOH is:

$pOH=-log(9.37x10^{-4})=3.03$

And the pH:

$pH=14-pOH=11.0$

In addition, the equilibrium concentrations of HClO and HF equals the change $x$ for each equilibrium as:

$[HClO]_{eq}=x_{ClO}=9.32x10^{-4}M$

$[HF]_{eq}=x_{F}=5.07x10^{-6}M$

Best regards.

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