Rate equations are expressed based on experimental data, which give
reaction. Whenever a reactant is in large excess, its concentration de
ed on experimental data, which give a clue to the mechanism of a
(a) For a reaction 2A + 2B - Products, rate of reaction is second order
marge excess, its concentration does not change appreciably with time.
Products, rate of reaction is second order with respect to A and first order with
respect to B. Represent the rate law.
(b) Rate of a reaction is given as r=k[A2B2C72. If concentration is expressed in mol L and time in
seconds, give the unit.of rate constant for the reaction.
(2)
(C) Rate of a reaction doubles on increasing the concentration of reactant by 8 times. Find the order of the
reaction.
(2)
(d) Find the order and molecularity of the reaction. A+ H2O(excess) → B+C
Answers
Answer:
Let the two prime numbers be 3 and 5.
Multiples of 3 and 5 are as follows
3 ... 3, 6, 9, 12, 15, 18,....
5 ... 5, 10, 15, 20, ....
LCM(3, 5) = 15
Since 3 and 5 are prime numbers.
HCF(3, 5) = 1
LCM(3, 5) × HCF(3, 5) = 15 × 1 = 15 ... (ii)
3 × 5 = 15 ...(iii)
From (ii) and (iii), we can say that
The product of any two prime number is equal to the product of their HCF and LCM.
Explanation:
Answer:
(A) The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders). For many reactions the rate is given by a power law such as.
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Explanation: