2 years ago

How does the number of reactants in a double-displacement reaction

Chemistry

1 answer:

mojhsa [17]2 years ago

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In a typical double displacement reaction, you would have a total of two products (AB + CD —> AD + BC).

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A decomposition reaction has a half life of 578 years. (a) What is the rate constant for this reaction? (b) How many years does

maw [93]

Answer:

$0.001199 year^{-1}$ is the rate constant for this reaction.

It will take $1.73\times 10^3 years$ to concentration to reach 12.5% of its original value.

Explanation:

A decomposition reaction follows first order kinetics:

Half life of the reaction = $t_{1/2}=578 years$

Rate constant of the reaction = k

For first order reaction, half life and rate constant are linked with an expression :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{578 years}=0.001199 year^{-1}$

$0.001199 year^{-1}$ is the rate constant for this reaction.

Initial concentration of reactant = $[A_o]$ = x

Final concentration of reactant after time t = $[A]$ = 12.5% of x = 0.125x

The integrated law of first order reaction :

$[A]=[A_o]\times e^{-kt}$

$0.125x=x\times e^{-0.001199 year^{-1}\times t}$

t = 1,734.31 years = $1.73\times 10^3 years$

It will take $1.73\times 10^3 years$ to concentration to reach 12.5% of its original value.

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