Question

The rate of disappearance of $S{O_2}$ in the reaction $2S{O_2} + {O_2} \to 2S{O_3}$ is $1.28 \times {10^{ - 5}}M{s^{ - 1}}$. The rate of appearance of $S{O_3}$ is
$
  A)0.64 \times {10^{ - 5}}M{s^{ - 1}} \\
  B)0.32 \times {10^{ - 5}}M{s^{ - 1}} \\
  C)2.56 \times {10^{ - 5}}M{s^{ - 1}} \\
  D)1.28 \times {10^{ - 5}}M{s^{ - 1}} \\
$

Answer 1

Hint: For reactants the rate of disappearance is a positive number. For products the rate of disappearance is a negative number because they are being formed and not disappearing. For reactants the rate of formation is a negative number because they are disappearing and not being formed.

Complete answer:
Reaction rate is calculated using the formula rate$ = \dfrac{{\Delta [C]}}{{\Delta T}}$, where $\Delta [C]$ is the change in product concentration during time period $\Delta T$. The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.
Rate of disappearance is given as $ - \dfrac{{\Delta [A]}}{{\Delta T}}$ where $A$ is a reactant. However, using this formula, the rate of disappearance cannot be negative. Also, if the negative rate of disappearance is essentially a positive rate of appearance.
The rate of appearance is a positive quantity. So, the rate of appearance of a product is equal to the rate of disappearance of a reactant.
So, for the reaction $2S{O_2} + {O_2} \to 2S{O_3}$ the stoichiometric ratio of $S{O_2}$ and $S{O_3}$ are same. So the rate will be the same.
Therefore, the rate of disappearance of $S{O_2}$ and the rate of formation of $S{O_3}$ are the same.
So, the correct answer is $D)1.28 \times {10^{ - 5}}M{s^{ - 1}}$

Note:
When a catalyst is involved in the collision between the reactant molecules, less energy is required for the chemical change to take place, and hence more collisions have sufficient energy for reaction to occur. The reaction rate therefore increases.