In 6 M HCl,the complex ion Ru(NH₃)₆³⁺ decomposes to a variety of products.The reaction is first order in Ru(NH₃)₆³⁺ and has a half-life of 14 hours at 25°C.Under these conditions,how long will it take for the [Ru(NH₃)₆³⁺] to decrease to 23.5% of its initial value?

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For a first-order reaction, the time it takes for a reactant to decrease to a certain percentage of its initial concentration can be calculated using the formula $t = \frac{\ln(\frac{[A]_0}{[A]})}{k}$, where $[A]_0$ is the initial concentration, $[A]$ is the final concentration, and $k$ is the rate constant. The half-life ($t_{1/2}$) of a first-order reaction is given by $t_{1/2} = \frac{0.693}{k}$. Given that the half-life is 14 hours, we can find the time it takes for the concentration to decrease to 23.5% of its initial value. Since 23.5% is less than half, we expect the time to be more than one half-life. The formula for calculating the time based on the half-life and the percentage remaining (in this case, 23.5% or 0.235 of the initial concentration) is $t = \frac{\ln(\frac{1}{0.235})}{\ln(2)} \times t_{1/2}$. Substituting the given half-life of 14 hours, we find that the time required is approximately 29 hours.

In 6 M HCl,the Complex Ion Ru(NH3)63+ Decomposes to a Variety

In 6 M HCl,the Complex Ion Ru(NH₃)₆³⁺ Decomposes to a Variety