Question

Concentrations of A, B and C vs time are graphed for the reaction $A⟶3B+C$. The rate of the reaction is defined as r = k [A] $(log2=0.3,log3=0.48)$
Find the correct options:

• A. $t_2=5t_1$
• B. $\left[A\right]$ at $t_2=1M$
• C. $C_1=3M$
• D. $t_{\frac{1}{2}}=20min$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $t_2=5t_1$
B $\left[A\right]$ at $t_2=1M$
C $C_1=3M$
D $t_{\frac{1}{2}}=20min$

$A⟶3B+C$
$t=0a_0$
$t=t{a}_0-x3xx$
When $\left[A\right]=\left(C\right)\Rightarrow x=\frac{a_0}{2}$
So, $t_{\frac{1}{2}}=20min$
When $\left[B\right]=\left[A\right]\Rightarrow a_0-x=3x\Rightarrow x=\frac{a_0}{4}$
So, $t_1=\frac{20}{log2}log\frac{4}{3}=\frac{20}{0.3}[0.6-0.48]$
$=8min$
$C_1=\frac{3a_0}{4}=3M$
When $\left[C\right]=3M\Rightarrow \left[A\right]=\frac{a_0}{4}So{t}_2=2t_{\frac{1}{2}}=40min;So{t}_2=5t_1$
And $\left[A\right]at{t}_2=1M$