Question

The progress of the reaction, $P \rightarrow n Q$ with time is represented in the graph shown below.


The initial rate of conversion of $P$ will be

• A. $0.2\, mol\, L ^{-1} h ^{-1}$
• B. $0.4\, mol\, L ^{-1} h ^{-1}$
• C. $0.8\, mol\, L ^{-1} h ^{-1}$
• D. $0.1\, mol\, L ^{-1} h ^{-1}$

Answer 1

Answer: (D)

From $t=0$ to $t=1\, h$ for $A$.
$[P]=0.6-0.5=0.1\, mol\, L ^{-1}$
$\therefore$ Initial rate of conversion of
$P =\frac{\Delta[A]}{\Delta t}=\frac{0.1\, mol\, L ^{-1}}{1\, h }$
$=0.1\, mol\, L ^{-1} h ^{-1}$