Question

Reaction, $5C⟶3A+2B$. The rate of formation of $A$ is $0.027mol{L}^{-1}se{c}^{-1}$. What will be the rate of disappearance of $C$?

• A. $0.090mol{L}^{-1}se{c}^{-1}$
• B. $0.027mol{L}^{-1}se{c}^{-1}$
• C. $0.063mol{L}^{-1}se{c}^{-1}$
• D. $0.045mol{L}^{-1}se{c}^{-1}$

Answer 1

Answer: (D)

$0.045mol{L}^{-1}se{c}^{-1}$

Rate of formation of $A$ is $0.027mol{L}^{-1}se{c}^{-1}$, which means $\frac{d\left[A\right]}{dT}$ is equal to $0.027mol{L}^{-1}se{c}^{-1}$.
Instantaneous rate of reaction in terms of $A$ can be written according to the equation as $\frac{1}{3}$$\frac{d\left[A\right]}{dt}$. Thus instantaneous rate is numerically equal to $\frac{1}{3}$$\times$$0.027$, which is $0.009$$mol{L}^{-1}se{c}^{-1}$.
Now according to the equation, instantaneous rate can also be written as $-$$\frac{1}{5}$$\frac{d\left[C\right]}{dt}$.
This means $-$$\frac{1}{5}$$\frac{d\left[C\right]}{dt}$ $=$ $0.009$.
Hence, $-$$\frac{d\left[C\right]}{dt}$$=$$0.045$.
Thus, rate of disappearance of $C$ is $0.045mol{L}^{-1}se{c}^{-1}$