Respuesta :
Explanation:
The given reaction equation is as follows.
[tex]2HI(g) \rightarrow H_{2}(g) + I_{2}(g)[/tex]
[tex]\frac{-d[HI]}{dt} = k[HI]^{2}[/tex]
[tex]-\int_{[HI]_{o}}^{[HI]_{t}} \frac{d[HI]}{[HI]^{2}} = k \int_{o}^{t} dt[/tex]
[tex]-[\frac{-1}{[HI]}]^{[HI]_{t}}_{[HI]_{o}} = kt[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{[HI]_{o}} = kt[/tex] .......... (1)
where, [tex][HI_{o}][/tex] = Initial concentration
[tex][HI]_{t}[/tex] = concentration at time t
k = rate constant
t = time
Now, we will calculate the initial concentration of HI as follows.
Initial rate = [tex]1.6 \times 10^{-7} mol/sec[/tex]
k = [tex]6.4 \times 10^{-9}[/tex]
R = [tex]k[HI]^{2}_{o}[/tex]
[tex][HI]^{2}_{o} = \frac{R}{k}[/tex]
= [tex]\frac{1.6 \times 10^{-7}}{6.4 \times 10^{-9}}[/tex]
[tex][HI]_{o}[/tex] = 5 M
Now, we will calculate the concentration of [tex][HI]_{t}[/tex] at t = [tex]1.01 \times 10^{10}[/tex] sec as follows.
Using equation (1) as follows.
k = [tex]6.4 \times 10^{-9}[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{5}[/tex] = [tex](6.4 \times 10^{-9}) \times 1.01 \times 10^{10}[/tex]
[tex]\frac{1}{[HI]_{t}}[/tex] = 64.44
[tex][HI]_{t}[/tex] = 0.0155 M
Thus, we can conclude that concentration of HI at t = [tex]1.01 \times 10^{10}[/tex] sec is 0.0155 M.
A reception subservient on the second and first-order reactants is called a second-order reaction.
The correct answer is:
The concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
The equation according to the question is:
2 HI (g) ⇒ H₂ (g) + I₂ (g)
[tex]\dfrac{\text{-d}\left[\begin{array}{ccc}\text{HI}\end{array}\right] }{\text{dt}} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2}[/tex]
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{t}} } - \dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{o}} } &= \text{kt}[/tex] .......equation (1)
Where, the initial concentration can be represented as: [tex]\left[\begin{array}{ccc}\text{HI}_{o} \end{array}\right][/tex]
The concentration at time t = [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex]
Rate constant will be = k
The time will be = t
The initial concentration of HI can be calculated as:
The initial rate = 1.6 × 10⁻⁷ mol/sec
k = 6.4 × 10⁻⁹
[tex]\text{R} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
[tex]\dfrac{\text{R}}{\text{k}} & =\left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
= [tex]\dfrac{1.6 \times 10^{-7} }{6.4 \times 10^{-9} }[/tex]
[tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right]\end{array}\right] _{0} &= 5 \;\text{M}[/tex]
To calculate the concentration [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex] at time (t) = 1.01 × 10 ¹⁰ sec.
Now, using the above equation: (1)
k = 6.4 × 10⁻⁹
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} - \dfrac{1}{5}[/tex]
= (6.4 × 10⁻⁹) × 1.01 × 10¹⁰
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 64.44[/tex]
[tex]{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 0.0155 \;\text{M}[/tex]
Therefore, concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
To learn more about second-order reaction refer to the link:
https://brainly.com/question/12446045
