For a given chemicall reaction
`gamma_(1)A+gamma_(2)Bto gamma_(3)C+gamma_(4)D`
Concentration of C changes from 10 mmol `dm^(-3)` to 20 mmol `dm^(-3)` in seconds .Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A . The rate of appearance of D has been ecperimentally determined to be `9 "mmol dm"^(-3) s^(-1)`. Therefore the rate of reaction is ________ mmol `dm^(-3) s^(-1)` .

Explanation

To solve the problem, we need to analyze the given chemical reaction and the information provided about the rates of change of concentrations.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Information:**
The reaction is given as:
\[ \gamma_1 A + \gamma_2 B \rightarrow \gamma_3 C + \gamma_4 D \]
We know:
- The concentration of C changes from \(10 \, \text{mmol dm}^{-3}\) to \(20 \, \text{mmol dm}^{-3}\).
- The rate of appearance of D is \(9 \, \text{mmol dm}^{-3} s^{-1}\).
- The rate of appearance of D is \(1.5\) times the rate of disappearance of B, which is twice the rate of disappearance of A.

2. **Calculate the Rate of Change of C:**
The change in concentration of C is:
\[ \Delta C = 20 \, \text{mmol dm}^{-3} - 10 \, \text{mmol dm}^{-3} = 10 \, \text{mmol dm}^{-3} \]
If this change occurs in \(t\) seconds, the rate of appearance of C is:
\[ \text{Rate of appearance of C} = \frac{\Delta C}{t} = \frac{10 \, \text{mmol dm}^{-3}}{t} \]

3. **Relate the Rates of Change:**
From the problem statement, we have:
- Rate of appearance of D (\(R_D\)) = \(9 \, \text{mmol dm}^{-3} s^{-1}\)
- Rate of disappearance of B (\(R_B\)) = \(\frac{R_D}{1.5} = \frac{9}{1.5} = 6 \, \text{mmol dm}^{-3} s^{-1}\)
- Rate of disappearance of A (\(R_A\)) = \(\frac{R_B}{2} = \frac{6}{2} = 3 \, \text{mmol dm}^{-3} s^{-1}\)

4. **Determine the Rate of Reaction:**
The rate of the reaction can be expressed in terms of the rates of disappearance of A, B, and the appearance of C and D. Using the stoichiometric coefficients:
\[ \text{Rate} = -\frac{1}{\gamma_1} \frac{d[A]}{dt} = -\frac{1}{\gamma_2} \frac{d[B]}{dt} = \frac{1}{\gamma_3} \frac{d[C]}{dt} = \frac{1}{\gamma_4} \frac{d[D]}{dt} \]
Assuming \(\gamma_1 = \gamma_2 = \gamma_3 = \gamma_4 = 1\) for simplicity, we can use:
\[ \text{Rate} = \frac{d[C]}{dt} = \frac{d[D]}{dt} = 9 \, \text{mmol dm}^{-3} s^{-1} \]

5. **Final Calculation:**
Since the rate of appearance of D is given as \(9 \, \text{mmol dm}^{-3} s^{-1}\) and we have established the relationships between the rates, we can conclude that the rate of the reaction is:
\[ \text{Rate of reaction} = 9 \, \text{mmol dm}^{-3} s^{-1} \]

### Final Answer:
The rate of reaction is \(9 \, \text{mmol dm}^{-3} s^{-1}\).