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The increase in reaction rate as a result of temperature rise from 10 K to 100 K is ...
512 increase of temperature $ n \times 10 $ Increase reaction rate = $ 2 ^ 9 $ $ \triangle T = 100 -10 = 90 = 9 \times 10 \therefore n = 9 $ $ \therefore Increases reaction rate = 2 ^ 9 = 512 $
At 300 K rate constant is $ 0.0231 min^{-1} $ , for a reaction. Bt at 320 K rate constant is $ 0.0693 min^ {-1} $ . The activation energy of the reaction is
$ 43.84 Kj mole ^ {-1} log { K_2 \over K_1} = { Ea \over 2.303 R } \left ( { T_2 -T_1 \over T_1 T_2 } \right) $ $ log \left ( { 0.0693 \over 0.0231 } \right) = { Ea \over 2.303 \times 8.3 } \left ( { 320 -300 \over 300 \times 320 } \right) $ $ log 3 = { Ea \over 1.901} \left( { 20 \over 96000} \right) $ Ea =43.84
The activation energy of a reaction is $ 9 Kcal mole^{-1} $ . The increase in the rate constant when its temperature is raised from 295 to 300 K is approximately
$ 1.289 times log { K_2 \over K_1 } = { Ea.DT \over 2.303 RT_2 T_1 } = { 9000 \times 5 \over 2.303 \times 2 \times 300 \times 295 } = 0.1104 $ $ log { K_2 \over K_1 } = 0.1104 , { K_2 \over K_1} = 1.289 , K_2 = K_1 \times 1.289 $
A reactant A forms two products. (i)$ A \rightarrow B \, activation energy E_1 $ (ii) $ A \rightarrow C \, activation energy E_2 $ $ If E_2 = 2 E_1 $ then $ K_1 and K_2 $ are related as
$ K_1 = K_2 A.e^{E_1 \over RT } , K_1 = A_1 .e^{-E_1 \over RT } , K_2 = A_2 .e^ {-E_2 \over RT } $
$ {K_1 \over K_2 } = { A_1 \over A_2 } \times e ^ { (E_2 - E_1 ) / RT } = A.e ^ { (2E_1 -E_1)/RT} = A.e^{E_1 / RT } $
$ \therefore K_1 = K_2 .A.e^{E_1 / RT } $
The activation energys of two reaction are $E_1 and E_2 (E_1 \gt E_2)$ . If the temperature of the system is increased from $ T_1 to T_2 $ , the rate constant of the reaction changes from $K_1 to K_2^1 $ in the first reaction and $ K_2 to K_2^1$ in second reaction, predict which of the following expression is correct ?
$ { K_1 ^ 1 \over K_1 } \gt { K_2 ^1 \over K_2 } $ $ log { K_1 ^1 \over K_1 } = { E_1 \over 2.303 R } \left[ { T_2 -T_1 \over T_1 T_2 } \right] , log { K_2^1 \over K_2 } = { E_2 \over 2.303 R } [ {T_2 - T_1 \over T_1 T_2 } ] $ $ Since E_1 \gt E_2 $ $ \therefore log { K_1 ^1 \over K_1 } \div log { K_2 ^1 \over K_2 } \gt 1 or { K_1^1 \over K_1 } \gt { K_2 ^1 \over K_2 }$
The rate of reaction $ 2x + y --> Products $ . $ Rate = K[x]^2[y] $ . If x is present in large excess, the order of the reaction is
1 The rate is not depend upon the reactant present in excess
$ CH_3 COOEt + H_2 O \rightarrow CHC_3 OOH + Et OH $ Order of reaction is .....
1
In which of the following cases, does the reaction go farthest to completion ?
∵ higher the equilibrium constant , fastest the rate of reaction to completion.
The activation energy of a reaction is zero. The rate constant of the reaction …
is nearly independent of temperature. $ Ea = 0 \therefore K = A.e ^ { -Ea /RT } = A.e ^ \circ = A $
Which of the following is the fast reaction ?
$ 6 CO_2 + ^ 6 H_2 O \rightarrow C_6 H_{12} O_6 + 6 O_2 $
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