Explanation
For the first order reaction, $$C_t=C_o\left (\cfrac {1}{2}\right)^n$$
where, $$n=\cfrac {t}{t_{1/2}}$$, $$\quad C_o=$$ Initial concentration
After $$3$$ half- lives of $$A$$, $${C_t}_{(A)}={C_o}_{(A)}\left(\cfrac {1}{2}\right)^\cfrac {3t_{1/2}}{t_{1/2}}=\cfrac {{C_o}_{(A)}}{8}$$
After $$3$$ half -lives of $$A$$, for $$B$$, $$\quad n=\cfrac {3\times 3}{4.5}=2$$
$$\therefore {C_t}_{(B)}={C_o}_{(B)}\left(\cfrac {1}{2}\right)^2=\cfrac {{C_o}_{(B)}}{4}$$
$$\therefore \cfrac {{C_t}_{(A)}}{{C_t}_{(B)}}=\left(\cfrac {{C_o}_{(A)}}{{C_o}_{(B)}}\right)\times \cfrac {4}{8}=\cfrac {1}{2}\times \cfrac {4}{8}=1:4$$
$$ \implies\Delta H = E_a(f) - E_a(b) $$ $$\implies -20 = 60 - E_a(b) $$ $$ \implies E_a(b) = 60 + 20 $$$$\implies 80 \space KJ mol^{-1}$$
Explanation:
From the above chart, we can conclude that after $$2$$ half-lives $$75$$% of reaction is completed.
For completion of $$87.5$$% reaction $$3$$ half lives are required.
For $$2$$ half lives $$100$$ $$mins$$ required.
Therefore, for $$1$$ half life $$50$$ $$mins$$ required.
For $$3$$ half lives $$3\times 50 $$ = $$150$$ $$mins$$ required.
Hence, the correct answer is option $$B$$.
The fraction of molecules having energy equal to or greater than $$E_a$$ is : $$ x = \dfrac{n}{N} = e^{-E_a/RT} (x = \frac{n}{N} = 0.0001\% = \frac{0.0001}{100} = 10^{-6})$$
$$ \therefore \log x = \dfrac{-E_a}{2.3 \times RT}$$ $$ \implies \ log 10^{-6} = \dfrac{-E_a}{2.3 \times 2 \times 400}$$ $$ \implies E_a = 6 \times 2.3 \times 800$$ $$ \implies E_a = 11.05 \space kcal/mol $$
Order = 1 (given)
$$ \because t_{75} = 2 \times t_{50} $$
$$ \implies 2 \times 15 = 30 $$
$$ k = \frac{0.693 }{t_{50}} = \frac{0.693}{15}$$
$$ a = 0.1 \space M$$
$$ (a-x) = 0.025 \space M$$
For first order reaction $$k = \frac{2.303}{t} \log(\frac{a}{a-x})$$
$$ \implies \frac{2.303 \times \log2}{15} = \frac{2.303}{t} \log(\frac{0.1}{0.025})$$
$$ \implies t= 30 \space minutes $$
At low pressure $$(P_A \rightarrow 0)$$, $$\theta_A$$ is very small and proportional to the pressure. The rate becomes first order rate (low P) = $$k_2K_AP_A$$
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