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According to the Arrhenius equation a straight line is to be obtained by plotting the logarithm of the rate constant of a reaction against ...
1 /T
The given reaction $ 2FeCl_3 + Sn Cl_2 \rightarrow 2 FeCl_2 + SnCl_4 $ is an example of ______ reaction
third order
In the reverable reaction $ 2 NO_2 \rightleftharpoons N_2 O_4 $ , the rate of disappearence of $ NO_2 $ is equal to ….
$ 2 K_1 [NO_2] ^ 2 - 2k_2 [N_2 O_4 ] $ $ For 2 NO_2 \rightleftharpoons N_2 O_4 $ $ Rate = - { 1 \over 2} { d [NO_2] \over dt} = K_1 [NO_2 ]^2 -K_2 [N_2 O_4 ] $ $ \therefore rate = { -d[NO_2 ] \over dt } = 2 K_1 [ NO_2 ]^2 - 2K_2 [N_2 O_4 ] $
If concentration of reactants is increased by ‘x’, then rate constant K becomes .
K
The rate constant is given by equation $ K = p.z.e^{-Ea/RT } $ which factor should register a decrease for the reaction to proceed more rapidly ?
E
For the reaction $ A + B \rightarrow C $ . the unit of rate constant is
$ sec^ {-1} mole ^ { -1} L $ $ \therefore Second order reaction $
The rate of the gaseous reaction is equal to K[A][B]. The volume of the vessel is suddenly reduced to one forth of the initial volume. The rate of reaction would be ...
16 / 1 Volume of the vessel is reduced to one foreth Concentration becomes 4 times
For reaction $ Y_2 + 2Z \rightarrow Product $ , rate controlling step is $ Y + ½ Z \rightarrow Q $ . If the concentration of Z is doubled, the rate of reaction will be
Becomes 1.414 times $ Rate = K [ Y ] [ Z ] ^ { 1 /2 } $ $ \therefore New rate =\sqrt 2 . k[Y] [Z] ^ {1 /2 } =1.414 K [Y] [Z] ^ {1/2 } $
The time for half lif of a certain reaction $ A \rightarrow Products $ , is one hour. When the initial concentration of the reactant A is $ 2 mol L^{-1} $ how much time does it take for its concentration to come from $ 0.50 to 0.25 mole L^{-1} $ if it is a zero order reaction ?
0.25 h For Zero order reaction $ K = { [A] _o \over 2 t { 1/ 2 }} = { 2 \over 2 \times 1 } = 1 mol L^ { -1} hr^{-1} $ $ t = { [ A]_o - [A] \over K } = { 0.50 - 0.25 \over 1 } = 0.25 hr $
For a first order reaction $ A \rightarrow Products $ , the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is
$ 3.47 \times 10 ^ {-4} M min^{-1} , K = { 2.303 \over 40 } log { 0.1 \over 0.025 } = 0.03466 min ^ {-1} $ $ Rate = K [A] ^1 = 0.03466 \times 0.01 =3.466 \times 10 ^ {-4} M min ^ {-1} $
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