Chemical Kinetics and Nuclear ChemistryHard
Question
Consider the following first-order decomposition reaction. A4(g)→ 4A(g)
Which of the following statement(s) is/are correct regarding the reaction? (log 2 = 0.3)
Options
A.At 30 min, only 20% reaction is complete.
B.t1/2 of reaction is 90 min.
C.Rate of reaction decreases linearly with time.
D.The time for intersection of two curves is independent of initial concentration of A4.
Solution
A4 ? 4A
t = 0 aM 0
t = 30 min (a – x) M 4xM
As $a - x = 4x \Rightarrow x = \frac{a}{5}$
∴ Percentage reaction at t = 30 min = $\frac{x}{a} \times 100 = 20\%$
Now, $30 = \frac{t_{1/2}}{\log 2}.\log\frac{a}{a - x} \Rightarrow t_{1/2} = 90\text{ min}$
Create a free account to view solution
View Solution FreeTopic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions
More Chemical Kinetics and Nuclear Chemistry Questions
The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnet in one direction, ...Rate law of the reaction A → Product is, rate = k[A]. Graphically it is represented as Hence, rate constant is :...Rates of hydration of the following alkenes are :-(P) CH3-O-CH = CH2(Q) F-CH = CH2(R) CH3 -CH = CH2(S) CH = CH2...According to kinetic theory of gases...The activation energies of two reactions are E1 and E2 with E1 > E2. If the temperature of the reacting system is inc...