For the reaction 2A + B `to` C, rate of disappearance of A 0.076 `mols^-1.`
What is the rate of formation of C

Answer

To solve the problem, we need to relate the rate of disappearance of reactants to the rate of formation of products using stoichiometry.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ 2A + B \rightarrow C \]

2. **Identify the relationship between the rates:**
From the balanced equation, we can express the rates of disappearance and formation as follows:
- The rate of disappearance of A is related to the rate of formation of C by the stoichiometric coefficients.
- For every 2 moles of A that disappear, 1 mole of C is formed.

3. **Express the rates mathematically:**
The rate of disappearance of A can be expressed as:
\[ -\frac{1}{2} \frac{d[A]}{dt} = \frac{d[C]}{dt} \]
Here, \(-\frac{d[A]}{dt}\) is the rate of disappearance of A, and \(\frac{d[C]}{dt}\) is the rate of formation of C.

4. **Substitute the given rate of disappearance of A:**
We know from the problem that the rate of disappearance of A is \(0.076 \, \text{mol s}^{-1}\). Therefore:
\[ -\frac{d[A]}{dt} = 0.076 \, \text{mol s}^{-1} \]

5. **Plug the value into the equation:**
Substituting the value into the rate equation gives:
\[ -\frac{1}{2} \cdot 0.076 = \frac{d[C]}{dt} \]

6. **Calculate the rate of formation of C:**
\[ \frac{d[C]}{dt} = -\frac{1}{2} \cdot 0.076 = 0.038 \, \text{mol s}^{-1} \]

### Final Answer:
The rate of formation of C is \(0.038 \, \text{mol s}^{-1}\).

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