When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is `10^(-6)` times, then activation energy of the reaction in the presence of enzyme is
(a)`(6)/(RT)`
(b)P is required
(c)Different forms `E_(a)` obtained in laboratory
(d)cannot say any things
When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is `10^(-6)` times, then activation energy of the reaction in the presence of enzyme is
(a)`(6)/(RT)`
(b)P is required
(c)Different forms `E_(a)` obtained in laboratory
(d)cannot say any things
(a)`(6)/(RT)`
(b)P is required
(c)Different forms `E_(a)` obtained in laboratory
(d)cannot say any things
A
6/RT
B
P is required
C
Different from Ea obtained in the laboratory
D
Cannot say anything
Text Solution
Verified by Experts
The correct Answer is:
C
Similar Questions
Explore conceptually related problems
For the first order reaction ArarrB+C , carried out at 27^(@)C if 3.8xx10^(-16)% of the reactant molecules exists in the activated state, the E_(a) (activation energy) of the reaction is :
Rate of a reaction increases by 10^6 times when a reaction is carried out in presence of enzyme catalyst at same temperature. Determine change in activation energy.
The rate of a certain biochemical reaction at physiological temperature (T) occurs 10^(6) times faster with enzyme than without. The change in the reaction energy upon adding enzyme is :
For a reaction the activation energy E_a=0 and the rate constant= 3.2 times 10^6 sec^-1 at 300K. What is the value of rate constant at 310K?
Read the following paragraph and answer the questions given below , Stable equilibrium is of various types . Mechanical equilibrium is achieved when all particles are at rest and total potential energy of the system is minimum . At any stage where particles are at rest but the system is not at stable equilibrium as it can reduce its potential energy by reverting to another position, is called metastable equilibrium . Thermal equilibrium is result from the absence of temperature gradients in the system. Chemical equilibrium is obtained when no further reaction occurs between reacting substances, i.e forward and reverse rates of reaction are equal. When steam with solid iron at high temperature Fe_3O_4(s) and hydrogen gas are produced . But the reaction never goes to completion. This is because as the products are formed the reaction proceeds in reverse direction and when rate of reverse reaction is equal to rate of forward reaction , the concentration of reactants and products become constant and equilibrium is reached. When two reactants (A) and (B) are mixed to give products (C) and (D) the concentration quotient (Q) at initial stage of the reaction
Read the following paragraph and answer the questions given below , Stable equilibrium is of various types . Mechanical equilibrium is achieved when all particles are at rest and total potential energy of the system is minimum . At any stage where particles are at rest but the system is not at stable equilibrium as it can reduce its potential energy by reverting to another position, is called metastable equilibrium . Thermal equilibrium is result from the absence of temperature gradients in the system. Chemical equilibrium is obtained when no further reaction occurs between reacting substances, i.e forward and reverse rates of reaction are equal. when steam with solid iron at high temperature Fe_3O_4(s) and hydrogen gas are produced . But the reaction never goes to completion. this is because as the products are formed the reaction proceeds in reverse direction and when rate of reverse reaction is equal to rate of forward reaction , the concentration of reactants and products become constant and equilibrium is reached. When a matchbox is balanced on its edge, it is unstable and will tumble to a more stable position resting on its face so as to acquire minimum potential energy. This is an example of
The beta - decay process , discovered around 1900 , is basically the decay of a neutron n . In the laboratory , a proton p and an electron e^(bar) are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. n rarr p + e^(bar) + bar nu _(e) , around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (bar nu_(e)) to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is 0.8 xx 10^(6) eV The kinetic energy carried by the proton is only the recoil energy. If the - neutrono had a mass of 3 eV// c^(2) (where c is the speed of light ) insend of zero mass , what should be the range of the kinectic energy K. of the electron ?
The beta -decay process, discovered around 1900 , is basically the decay of a neutron (n) , In the laboratory, a proton (p) and an electron (e^(-)) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a tro-body dcay process, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., n rarr p + e^(-)+overset(-)v_(e ) , around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (overset(-)V_(e )) to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8xx10^(6)eV . The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino has a mass of 3eV//c^(2) (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K of the electron?
It we see the reaction of methane with halogen, the rate determining step for chlorination is, endothermic reaction of the chlorine atom with methane to form methyl radical and a molecule of HCl. So free radical is the intermediate of the reaction. Formation of free radical depends upon the energy required to break a bond between a hydrogen atom and a carbon atom. Chlorination of propane and Bromination of propane. when compared it is found that bromination is more selective than chlorination. The probability factor for 3^(@),2^(@),1^(@)H atom is 5.0:3.8:1.0 at 25^(@)C for chlorination. Isobutane when reacts with chlorine in presence of ultra violet radiations yield 2 products primary hydrogen substituted and 3^(@) hydrogen substituted Find their % in product mixture
The reaction 2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g) has been studied kinetically and on the baiss of the rate law following mechanism has been proposed. I. 2A X hArr A_(2)X_(2) " " ("fast and reverse") II. A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X III. A_(2)X+B_(2)rarrA_(2)+B_(2)X where all the reaction intermediates are gases under ordinary condition. form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step. Let the equilibrium constant of Step I be 2xx10^(-3) mol^(-1) L and the rate constants for the formation of A_(2)X and A_(2) in Step II and III are 3.0xx10^(-2) mol^(-1) L min^(-1) and 1xx10^(3) mol^(-1) L min^(-1) (all data at 25^(@)C) , then what is the overall rate constant (mol^(-2) L^(2) min^(-1)) of the consumption of B_(2) ?
Recommended Questions
- When a biochemical reaction is carried out in laboratory from outside ...
Text Solution
|
- When a biochemical reaction is carried out in laboratory from outside ...
Text Solution
|
- Enzymes catalyse the biochemical reactions by the activation energy.
Text Solution
|
- The rate of a certain biochemical reaction c arried by enzymes in huma...
Text Solution
|
- Rate of a reaction increases by 10^6 times when a reaction is carried ...
Text Solution
|
- When a biochemical reaction is carried out in laboratory from outside ...
Text Solution
|
- When a biochemical reaction is carried out in laboratory from outside ...
Text Solution
|
- जब प्रयोगशाला में जैव रासायनिक अभिक्रिया एन्जाइम की अनुपस्थिति में करा...
Text Solution
|
- जब प्रयोगशाला में जैव रासायनिक अभिक्रिया एन्जाइम की अनुपस्थिति में करा...
Text Solution
|