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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} $
In the reaction $ 2N_2O_5 \rightarrow 4NO_2 + O_2 , initial pressure is 500 atm and rate constant K is 3.38 10^{-5} sec^{-1} . After 10 minutes the final pressure of N_2O_5 $ is
490 atm , $ K = { 2.303 \over t} log { Po \over Pt } \therefore 3.38 \times 10 ^ {-5} = { 2.303 \over 600 } log { 500 \over Pt } $ $ log { 500 \over Pt } = 0.0088 OR { 500 \over pt } = 1.021 OR pt = 490 atm $
The rate constants $K_1 and K_2 for two different reactions are 10^{16}.e^{-2000/T} and 10^{15} .e^{-1000/T} $ respectively. The temperature at which $ K_1 = K_2 $ is
$ { 1000 \over 2.303 } k , k_1 = k_2 $ $ \therefore 10 ^ {16} .e^{-2000 / T } = 10 ^ {15 } .e^{-1000 / T } $ $ \therefore 10.e^{-2000 / T } = 1 .e ^ {-1000 / T } $ $ \therefore ln 10 - {2000 \over T } = - { 1000 \over T } $ $ \therefore 2.303 - { 2000 \over T } = - {1000 \over T } $ $ \therefore T = - { 1000 \over 2.303 } K $
The temperature of the system decreases in an
In an adiabatic process, there is no heat exchange with the surroundings. During adiabatic expansion, the system does work on the surroundings, and since no heat is added to the system, its internal energy decreases, resulting in a decrease in temperature.
If a refrigerator’s door is opened, then we get
When a refrigerator door is opened, the cooling system works harder to cool the extra warm air entering from the room. The heat extracted from the refrigerator's interior is released into the room, resulting in an overall increase in the room's temperature.
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