Upon dissolving in, undergoes a disproportionation reaction according to the following unbalanced equation: This disproportionation follows first-order kinetics with a half-life of s. What is the concentration of after h if the initial solution of was prepared by dissolving in dilute to make of solution? What mass of is formed after h?

The question in the narrative seems not to be complete because the unbalanced equation was not given and the values of the second (s) and chemical dissolved was also not given.

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Correct Question:

Upon dissolving InCl (s) in HCl, In (aq) undergoes a disproportionation reaction according to the following unbalanced equation:

In ⁺ (aq) → In (s) → In³⁺ (aq)

This disproportionation follows first-order kinetics with a half-life of 667s. What is the concentration of In⁺ (aq) after 1.25 h if the initial solution of In⁺ (aq) was prepared by dissolving 2.38 g InCl (s) in dilute HCl to make 5.00 x 10² mL of solution? What mass of In (s) is formed after 1.25h?

Solution / Explanation:

Given half life of In⁺ at 66.7 s,

We recall the formula used in claculating the rate of constant for the first oreder reactin as:

K = 0.693 / t₁:2,

Noting that:

t₁:2 = half life

and K = rate constant

Therefore, if we replace the value of t₁:2 in the formular above,

We have,

K = 0.693 / 667s

K = 0.00 / s

Now, if we recall the mass of InCl (s) as 2.38g,

Volume of dilute HCl = 500 mL,

and the molar mass of InCl (s) - 150.271 g/mol,

The number of moles is then calculated using the formular:

Number of moles: = Given Mass/Molar Mass

Now replacing the given values of given mass and the molar mass in the above formular,

= 2.38g / 150.271g/mol

= 0.0158 mol

Volume of diluted 500 mL.

Recalling also that we need to convert from mL into Liters

Therefore, 1mL = 10⁻³L

Therefore,

500mL = (500 X 10⁻³) L

0.5 L

Now, the molarrity of In⁺ (aq) is calculated using

morality of In⁺ (aq) = moles of In⁺ (aq) / volume of solution

= 0.0158/0.5L

=0.0316M (This is the initial concentration of In⁺ (aq))

The time of the reaction is 1.25h

There is 3600s in one hour

1.25h = 1.25 x 3600

= 4500s