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Which of the following represents the expression for 3/4 th life of a first order reaction?
t3/4 = (2.303/K) log (a/(a/4)]
t3/4 = (2.303/K)log4
For an endothermic reaction, where ΔH represents the enthalpy of the reaction in kJ/mol, the minimum value for the energy of activation will be
For endothermic reaction, energy of activation is always greater than H.
In the formation or sulphur trioxide by the contact process,
2SO2(g) + O2(g) 2SO3(g)
The rate of reaction is expressed as d(O2)/dt = 2.5x10-4 mol L-1sec-1. The rate disappearance of (SO2) will be
2SO2 + O2 2SO3(g)
-1/2dsO2/dt = -dO2/dt
-dsO2/dt = 2x2.5x10-4 =5x10-4 mol L-1sec-1
A first order reaction complete 40% in 20 minutes. In how much time it is completed 90% of its original amount?
K = (2.303/t)log(a/a-x)
when x = (40/100)a = 0.4a
t = 20min
K = (2.303/20)log(a/a-0.4a)
=(2.303/20)log1/0.6
K = 0.02554 min-1
t = (2.303/0.02554)log(a/a-0.9a)
=2.303/0.02554 log (1/0.1)
=90.17 min
The dissociation of H2O2 is a first order reaction. The half life for '16 V H2O2 is 30 min, calculate the time at which the solution is 1V H2O2 ?
t = 2.303/(0.693/30)log 24/1
or, t = 30x4xlog2/0.3010
or, t =120min.
How fast the reaction is at 25°C as compared to 0°C. If the activation energy is 65 KJ :-
The rate constant of a reaction depends on temperature according to the Arrhenius equation: k = A * e^(-Ea/RT). An increase in temperature from 0°C to 25°C raises the rate by a factor of e^(65000/8.314*(1/273 - 1/298)) ≈ 11 times, for an activation energy of 65 kJ/mol.
When temperature is increased from 27C to 127C, rate of reaction becomes doubled, then Ea will be?
For a certain reaction, 10% of the reactant dissociates in 1 hour, 20% of the reactant dissociate in 2 hour, 30% of the reactant dissociates in 3 hour. Then the units of the rate constant is:-
The units of rate constant for a first-order reaction are mol Lâ»Â¹ hrâ»Â¹ or sâ»Â¹. This is because the rate is proportional to the concentration of the reactant, which has units of mol Lâ»Â¹, and the rate constant has units of timeâ»Â¹, such as hrâ»Â¹ or sâ»Â¹.
The activation energy of a reaction is zero. The rate constant of the reaction is :
K = Ae-Ea/RT
K= Ae0/RT
K = A= const.
The rate constant of a first order reaction is 10-3 min-1 at 27°C. The temperature coefficient of this reaction is 2. The rate constant at 17°C will be :
KT1/KT2 =T/10= 10-3/KT1=10/10 = 2
KT1 = 10-3/2 = 0.5 x 10-3 =5x 10-4 min-1
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