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For an endothermic reaction $ A --> B . An activation energy of 15 Kcal mole^{-1} and the enthalpy change of reaction is 5 Kcal mole^{-1} . The activation energy for the reaction B --> A $ is
10 Kcal $ mole^{-1} \triangle H = Ea - Ea^r \therefore +5 = 15 - Ea ^5 \therefore Ea^r = 10 $
For an exothermic reaction an activation energy of $ 70 KJ mole^{-1} $ and the enthalpy change of reaction is $ 30 KJ mole^{-1} $ . The activation energy for the reverse reaction is ...
100 Kj $ mole ^ {-1} \triangle H = Ea - Ea^r -30 = 70 - Ea^r \therefore Ea^r = 100 $
The rate constant of the reaction increases by ...
increasing the temperature.
Which of the following is the expression for Arrhenius equation ?
All the above
The Plot of log $ K vs {1 \over T } $ helps to calculate
Activation energy and frequency factor.
At 290 K velocity constant of a reaction was found to be $ 3.2 \times 10^ {-3} $ . At 300 K, it will be
$ 3.2 \times 10^{–4} 10k $ rise, the velocity constant becomes nearly double
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 } $
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