JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 14)
For a first order reaction, A $$ \to $$ P, t1/2 (half-life) is 10 days The time required for $${1 \over 4}$$th conversion of A (in days) is : (ln 2 = 0.693, ln 3 = 1.1)
5
3.2
4.1
2.5
Explanation
For first order reaction,
The half life, t$${1 \over 2}$$ = $${{0.693} \over k}$$
Here given t$${1 \over 2}$$ = 10 days
$$\therefore\,\,\,$$ k = $${{0.693} \over {10}}$$ = 0.0693 days$$-$$1
Now, the time required for $${1 \over 4}$$th conversion of A is ,
t = $${{2.303} \over k}$$ log10 $$\left( {{a \over {a - x}}} \right)$$
= $${{2.303} \over {0.0693}}$$ log $$\left( {{1 \over {1 - {1 \over 4}}}} \right)$$
= $${{2.303} \over {0.0693}}$$ log $$\left( {{4 \over 3}} \right)$$
= 4.1 days.
The half life, t$${1 \over 2}$$ = $${{0.693} \over k}$$
Here given t$${1 \over 2}$$ = 10 days
$$\therefore\,\,\,$$ k = $${{0.693} \over {10}}$$ = 0.0693 days$$-$$1
Now, the time required for $${1 \over 4}$$th conversion of A is ,
t = $${{2.303} \over k}$$ log10 $$\left( {{a \over {a - x}}} \right)$$
= $${{2.303} \over {0.0693}}$$ log $$\left( {{1 \over {1 - {1 \over 4}}}} \right)$$
= $${{2.303} \over {0.0693}}$$ log $$\left( {{4 \over 3}} \right)$$
= 4.1 days.
Comments (0)
