For a reaction `pA + qB rarr` Product, the rate law expresison is `r = k[A][B]^(m)`. Then

Options

A. `(p+q) = (1+m)`
B. `(p+q)gt(1+m)`
C. `(p+q)` may or may not b equal to `(1+m)`
D. `(p+q) = (1+m)`

Answer

To solve the problem, we need to analyze the relationship between the stoichiometric coefficients of the reactants and the orders of the reaction as given in the rate law expression.

### Step-by-Step Solution:

1. **Identify the Reaction and Rate Law**:
- The reaction is given as \( pA + qB \rightarrow \text{Products} \).
- The rate law expression is provided as \( r = k[A]^1[B]^m \).

2. **Understand Stoichiometric Coefficients**:
- In the reaction, \( p \) and \( q \) are the stoichiometric coefficients for reactants \( A \) and \( B \) respectively.
- Stoichiometric coefficients indicate the number of moles of each reactant involved in the reaction.

3. **Understand Reaction Orders**:
- The rate law indicates the order of the reaction with respect to each reactant.
- The order with respect to \( A \) is 1 (since the exponent of \( [A] \) is 1).
- The order with respect to \( B \) is \( m \) (since the exponent of \( [B] \) is \( m \)).

4. **Establish the Relationship**:
- We need to find the relationship between \( p + q \) and \( 1 + m \).
- For elementary reactions, the stoichiometric coefficients are equal to the reaction orders. Thus, if the reaction is elementary, we would have:
\[ p + q = 1 + m \]
- However, the problem does not specify that the reaction is elementary.

5. **Conclusion**:
- Since the reaction may not be elementary, we cannot definitively state that \( p + q \) is equal to \( 1 + m \). Therefore, the relationship can be expressed as:
\[ p + q \text{ may or may not be equal to } 1 + m \]

### Final Answer:
The relationship between \( p + q \) and \( 1 + m \) is that \( p + q \) may or may not be equal to \( 1 + m \).

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