Question

$2C$$+$$O_2$$=$$2CO$. The rate of disappearance of $C$ is $2\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$. What is the instantaneous rate of the reaction?

• A. $2\times {10}^{-3}$
• B. $1\times {10}^{-3}$
• C. $4\times {10}^{-3}$
• D. None of these

Answer 1

The correct option is B $1\times {10}^{-3}$

The rate of disappearance of $C$ is $2\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$, which means $\frac{d\left[C\right]}{dt}$ is $2\ast {10}^{-3}mol{L}^{-1}se{c}^{-1}$. Now, instantaneous rate of reaction in terms of reactant can be written as $-$$\frac{1}{2}$$\frac{d\left[C\right]}{dt}$.
Thus, $-$$\frac{1}{2}$$\frac{d\left[C\right]}{dt}$$=$$-$$\frac{1}{2}$$\times$$2\times {10}^{-3}$, which is equal to $-$$1\times {10}^{-3}mol{L}^{-1}se{c}^{-1}$. The negative sign is just an indication that the concentration of the reactant decreases over time. Hence, $1\times {10}^{-3}$ is the answer.