Question

$A$$+$$2B$$=$$4C$. The rate of appearance of $C$ at a time $t$ is $1.5\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$. What will be the instantaneous rate of the reaction?

• A. $6.0\times {10}^{-3}$
• B. $1.5\times {10}^{-3}$
• C. $0.675\times {10}^{-3}$
• D. $0.375\times {10}^{-3}$

Answer 1

The correct option is D $0.375\times {10}^{-3}$

The rate of appearance of $C$ at a time $t$ is $1.5\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$, which means $\frac{d\left[C\right]}{dt}$ is $1.5\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$. Now instantaneous rate of the reaction in terms of product can be written as $\frac{1}{4}$$\frac{d\left[C\right]}{dt}$.
Thus, $\frac{1}{4}$$\frac{d\left[C\right]}{dt}$ $=$ $\frac{1}{4}$$\times$$1.5\times {10}^{-3}$, that is $0.375\times {10}^{-3}$.
Hence, instantaneous rate of the reaction is $0.375\times {10}^{-3}$$mol{L}^{-1}se{c}^{-1}$.