The progress of the reaction, P arrow n Q with time is represented in the graph shown below. The initial rate of conversion of P will be

Question

The progress of the reaction, P arrow n Q with time is represented in the graph shown below. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/e53e3b55debc91a09154f250074bbc67-.png" /> 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

Tardigrade Admin · Accepted Answer

From t=0 to t=1 h for A. [P]=0.6-0.5=0.1 mol L -1 ∴ Initial rate of conversion of P =(Δ[A]/Δ t)=(0.1 mol L -1/1 h ) =0.1 mol L -1 h -1

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

The initial rate of conversion of $P$ will be

$0.2 m o l L^{- 1} h^{- 1}$

$0.4 m o l L^{- 1} h^{- 1}$

$0.8 m o l L^{- 1} h^{- 1}$

$0.1 m o l L^{- 1} h^{- 1}$

From $t = 0$ to $t = 1 h$ for $A$ .
$[P] = 0.6 - 0.5 = 0.1 m o l L^{- 1}$
$∴$ Initial rate of conversion of
$P = \frac{Δ [ A ]}{Δ t} = \frac{0.1 m o l L ^{- 1}}{1 h}$
$= 0.1 m o l L^{- 1} h^{- 1}$

Q. The progress of the reaction, P→nQ with time is represented in the graph shown below. The initial rate of conversion of P will be

0.2molL−1h−1

0.4molL−1h−1

0.8molL−1h−1

0.1molL−1h−1