InterviewSolution
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InterviewSolution
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Two reaction, `(I)A rarr` Products and `(II) B rarr` Products, follow first order kinetics. The rate of reaction `(I)` is doubled when the temperature is raised form `300 K` to `310K`. The half life for this reaction at `310K` is `30 min`. At the same temperature `B` decomposes twice as fast as `A`. If the energy of activation for reaction `(II)` is twice that of reaction `(I)`, (a) calculate the rate of constant of reaction `(II)` at `300 K`. |
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Answer» Correct Answer - `0.03267 min^(-1)` For reaction `(I)` `A rarr` Products `T_(1) = 300 K, T_(2) = 310 K` `(k_(1))_(A)` i s at `300 K` and `(k_(2))_(A)` is at `310 K` Given: `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = 2` …(i) Uisng Arrhenius equation for reaction `(I)` : `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` ...(ii) `log 2 = (E_(a(A)))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iii) (For reaction `II`) `B rarr` Products Since at `310 K , B` decomposes twice as fast as `A`. `because [k_(2)(310 K)]_(A) = 0.0231 min^(-1)` `:. [k_(2)(310 K)]_(B) = 2 xx 0.0231 min^(-1)` `= 0.0462 min^(-1)` We have to calculate `[k_(1) (300 K)]_(B) =` ? Uisng Arrhenius equation for reaction `(II)` `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (E_(a(B)))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iv) Given: `(E_(a))_(B) = (1)/(2)(E_(a))_(A)` ...(v) Substitute the value of Eq. (v) in Eq. (iv), `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` (vi) Comparing Eqs. (iii) and (vi), we get `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx log 2` `:. (0.0462)/([k_(1)(300 K)]_(B)) = (2)^((1)/(2)) = sqrt(2) = 1.414` ....(vii) `:. [k_(1) (300 K)]_(B) = (0.0462)/(1.414) = 0.03267 min^(-1)` Correct Answer - `0.03267 min^(-1)` For reaction `(I)` `A rarr` Products `T_(1) = 300 K, T_(2) = 310 K` `(k_(1))_(A)` i s at `300 K` and `(k_(2))_(A)` is at `310 K` Given: `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = 2` …(i) Uisng Arrhenius equation for reaction `(I)` : `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` ...(ii) `log 2 = (E_(a(A)))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iii) (For reaction `II`) `B rarr` Products Since at `310 K , B` decomposes twice as fast as `A`. `because [k_(2)(310 K)]_(A) = 0.0231 min^(-1)` `:. [k_(2)(310 K)]_(B) = 2 xx 0.0231 min^(-1)` `= 0.0462 min^(-1)` We have to calculate `[k_(1) (300 K)]_(B) =` ? Uisng Arrhenius equation for reaction `(II)` `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (E_(a(B)))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iv) Given: `(E_(a))_(B) = (1)/(2)(E_(a))_(A)` ...(v) Substitute the value of Eq. (v) in Eq. (iv), `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` (vi) Comparing Eqs. (iii) and (vi), we get `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx log 2` `:. (0.0462)/([k_(1)(300 K)]_(B)) = (2)^((1)/(2)) = sqrt(2) = 1.414` ....(vii) `:. [k_(1) (300 K)]_(B) = (0.0462)/(1.414) = 0.03267 min^(-1)` |
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