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Yuki888 [10]
3 years ago
5

If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain afte

r 10.0 minutes?
Chemistry
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

m=0.127g

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g

Best regards.

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Explanation:

The reaction quotient Qc is defined for a generic reaction:

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where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

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This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

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  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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