Advertisements
Advertisements
प्रश्न
Assertion: The enthalpy of reaction remains constant in the presence of a catalyst.
Reason: A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
पर्याय
Both assertion and reason are correct and the reason is correct explanation of assertion.
Both assertion and reason are correct but reason does not explain assertion.
Assertion is correct but reason is incorrect.
Both assertion and reason are incorrect.
Assertion is incorrect but reason is correct.
Advertisements
उत्तर
Both assertion and reason are correct and the reason is correct explanation of assertion.
Explanation:
AH = Activation Energy of forward reaction – Activation Energy of reverse reaction.
Catalyst does not alter heat of reaction because it affects activation energy of forward and reverse reactions equally.
APPEARS IN
संबंधित प्रश्न
In a first order reaction x → y, 40% of the given sample of compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
A → B is a first order reaction with rate 6.6 × 10-5m-s-1. When [A] is 0.6m, rate constant of the reaction is
- 1.1 × 10-5s-1
- 1.1 × 10-4s-1
- 9 × 10-5s-1
- 9 × 10-4s-1
Write molecularity of the following reaction:
2NO(g)+O2(g)→2NO2(g)
Write two factors that affect the rate of reaction.
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{CH3CHO_{(g)} -> CH4_{(g)} + CO_{(g)}}\] Rate = k[CH3CHO]3/2
For the reaction:
\[\ce{2A + B -> A2B}\]
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
A reaction is first order in A and second order in B. Write the differential rate equation.
A reaction is first order in A and second order in B. How is the rate affected when the concentrations of both A and B are doubled?
The decomposition of N2O5(g) at 320K according to the following equation follows first order reaction:
`N_2O_(5(g))->2NO_(2(g))+1/2O_(2(g))`
The initial concentration of N2O5(g) is 1.24 x 10-2 mol. L-1 and after 60 minutes 0.20x10-2 molL-1. Calculate the rate constant of the reaction at 320K.
Molecularity of a reaction _____________.
Which of the following statement is true for order of a reaction?
Consider a first order gas phase decomposition reaction given below :
\[\ce{A(g) -> B(g) + C(g)}\]
The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’ The rate constant k for the reaction is given as ______.
In any unimolecular reaction:
(i) only one reacting species is involved in the rate determining step.
(ii) the order and the molecularity of slowest step are equal to one.
(iii) the molecularity of the reaction is one and order is zero.
(iv) both molecularity and order of the reaction are one.
Assertion: Order of the reaction can be zero or fractional.
Reason: We cannot determine order from balanced chemical equation.
Assertion: Order and molecularity are same.
Reason: Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.
Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.
On heating compound (A) gives a gas (B) which is constituent of air. The gas when treated with H2 in the presence of catalyst gives another gas (C) which is basic in nature, (A) should not be ______.
For a chemical reaction starting with some initial concentration of reactant At as a function of time (t) is given by the equation,
`1/("A"_"t"^4) = 2 + 1.5 xx 10^-3` t
The rate of disappearance of [A] is ____ × 10-2 M/sec when [A] = 2 M.
[Given: [At] in M and t in sec.]
[Express your answer in terms of 10-2 M /s]
[Round off your answer if required]
Higher yield of NO in \[\ce{N2(g) + O2 <=> 2NO(g)}\] can be obtained at:
[ΔH of the reaction = +180.7 kJ mol−1]
- higher temperature
- lower temperature
- higher concentration of N2
- higher concentration of O2
Choose the correct answer from the options given below:
