Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI and the rate constant is 9.7×10−6M−1s−1. If the initial concentration of HI is 0.140 M .

What is its molarity after a reaction time of 7.00 days?

What is the time (in days) when the HI concentration reaches a value of 8.0×10−2 M ?

Answer :

Answer:

For 1: The concentration after the given time is 0.077 M

For 2: The time taken to reach the given value is 6.39 days

Explanation:

The integrated rate law equation for second order reaction follows:

[tex]k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)[/tex]      ......(1)

where,

k = rate constant = [tex]9.7\times 10^{-6}M^{-1}s^{-1}[/tex]

t = time taken

[tex][A][/tex] = concentration of substance after time 't'

[tex][A]_o[/tex] = Initial concentration = 0.140 M

  • For 1:

We are given:

t = 7.00 days = (7 × 24 × 60 × 60) = 604800 seconds   (1 day = 24 hours, 1 hour = 60 mins, 1 min = 60 sec)

[A] = ?

Putting values in equation 1, we get:

[tex]9.7\times 10^{-6}=\frac{1}{604800}\left (\frac{1}{[A]}-\frac{1}{0.140}\right)[/tex]

[tex][A]=0.077M[/tex]

Hence, the concentration after the given time is 0.077 M

  • For 2:

We are given:

[A] = [tex]8.0\times 10^{-2}M[/tex]

Putting values in equation 1, we get:

[tex]9.7\times 10^{-6}=\frac{1}{t}\left (\frac{1}{(8.0\times 10^{-2})}-\frac{1}{0.140}\right)[/tex]

[tex]t=552283s=6.39days[/tex]

Hence, the time taken to reach the given value is 6.39 days

The time when the HI concentration reaches a value of 8.0 × 10⁻² M will be "6.39 days".

Hydrogen Iodide

According to the question,

Rate constant, k = 9.7 × 10⁻⁶ H⁻¹ s⁻¹ or,

                           = 9.7 × 10⁻⁶ L/mol.s

Concentration of HI = 0.140 M

We know that,

→ [tex]\frac{1}{[HI]}[/tex] = kt + [tex]\frac{1}{[HI]_0}[/tex]

Now,

  [HI] = [tex][kt + \frac{1}{[HI]_0} ]^{-1}[/tex]

By substituting the values, we get

        = (9.7 × 10⁻⁶ × [tex]\frac{60}{1}[/tex] × [tex]\frac{60}{1}[/tex] × [tex]\frac{24}{1}[/tex] × 7.00 + [tex]\frac{1}{0.140}[/tex])⁻¹

        = (5866560 × 10⁻⁶ + [tex]\frac{1}{0.140}[/tex])⁻¹  

        = ([tex]\frac{(0.140\times 5.866560)(1\times 1)}{0.140}[/tex])⁻¹

        = (13.00941714)⁻¹ or,

        = 0.0786 mol/L (Molarity of [HI])

Now,

[HI] = [tex][kt + \frac{1}{[HI]_0} ]^{-1}[/tex]

   t = [tex]\frac{1}{k}[/tex] ([tex]\frac{1}{[HI]} -\frac{1}{[HI]_0}[/tex])

By substituting the values,

     = [tex]\frac{1}{9.7\times 10^{-6}}\times \frac{1}{60}\times \frac{1}{60}\times \frac{1}{24}[/tex] ([tex]\frac{1}{0.080} -\frac{1}{0.140}[/tex])

     = [tex]\frac{1}{838080\times 10^{-6}}[/tex] × ([tex]\frac{0.06}{0.0112}[/tex])

     = 6.39 days

Thus the above answer is correct.

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