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For the reaction:
2N2O5(g) → 4NO2(g) + O2(g) the rate law is: (Δ[O2]/Δt) = k[N2O5] At 300 K, the half-life is 2.50 × 104 seconds and the activation energy is 103.3 kJ/mol. What is the half-life at 310 K? (Hint: Use rate law expression to determine the reaction order → solve for k1 at 300 K using the corresponding half-life expression → use two-point Arrhenius equation to solve for k2 at 310 K → use the half-life expression again to solve for half-life at 310 K)

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dsdrajlin

Answer:

Half-life at 310 K is 6.54 × 10³ s.

Explanation:

Let's consider the following reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

The rate law is:

(Δ[O₂]/Δt) = k . [N₂O₅]

Since [N₂O₅] is raised to the power of 1, the reaction order is 1.

For a first-order reaction:

[tex]t_{1/2}=\frac{ln2}{k}[/tex]

where,

[tex]t_{1/2}[/tex] is the half-life

[tex]k[/tex] is the rate constant

At 300 K,

[tex](t_{1/2})_{1}=\frac{ln2}{k_{1}}\\k_{1}=\frac{ln2}{(t_{1/2})_{1}} =\frac{ln2}{2.50 \times 10^{4} s} =2.77 \times 10^{-5} s^{-1}[/tex]

We can use two-point Arrhenius equation to solve for k₂ at 310 K

[tex]ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} (\frac{1}{T_{2}} -\frac{1}{T_{1}} )\\ln\frac{k_{2}}{k_{1}} =\frac{-103.3kJ/mol}{8.314 \times 10^{-3} kJ/mol.K} .(\frac{1}{310K}-\frac{1}{300K}  )\\ln\frac{k_{2}}{k_{1}}=1.34\\k_{2}=1.06 \times 10^{-4} s^{-1}[/tex]

At 310 K,

[tex](t_{1/2})_{2}=\frac{ln2}{k_{2}}=\frac{ln2}{1.06 \times 10^{-4} s^{-1}   } =6.54 \times 10^{3} s[/tex]

The study of chemicals is called chemistry. When the reactant reaches the activation energy it will convert to the products in any case. The energy required to reach the activation energy is called threshold energy.

The correct answer is [tex]6.54*10^{3}[/tex].

The reaction is mentioned as follows:-

[tex]2 N_2O_5(g) ---> 4 NO_2(g) + O_2(g[/tex])

Activation energy

  • The minimum amount of energy required to reach the threshold is called activation energy.

The rate law is for the equation is as follows:-

[tex]\frac{([O_2]}{Δt} = k . [N_2O_5][/tex]

  • Since [tex][N_2O_5][/tex]reaction order is 1. The formula used for the first-order reaction is as follows:-

[tex]T_\frac{1}{2} = \frac{In_2}{K}[/tex] where,

  • T is the half-life
  • K is constant

At 300K the value of T half is [tex]2.77*10^{-5}[/tex]

  • Use the Arrhenius equation to solve the question.

[tex]In\frac{K_2}{K_1} =\frac{-Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]

Place all the value in the question and solve it

[tex]In\frac{K_2}{K_1} =\frac{-103.3}{8.314*10^{-3}}(\frac{1}{310}-\frac{1}{300})\\\\= 1.06*10^{-4}[/tex]

Hence, the correct answer is [tex]1.06*10^{-4}[/tex].

For more information about the question, refer to the link.

https://brainly.com/question/491373

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