For a hypothetical reaction `A + 2B to 3C` Identify x, y and z.
`-1/x (dA)/(dt) = -1/y (dB)/(dt) =1/z (dC)/(dt)`

Options

A. 1,2,3
B. 0,2,1
C. 3,2,1
D. 1,1,2

Answer

To solve the problem, we need to analyze the given hypothetical reaction:

\[ A + 2B \rightarrow 3C \]

We are provided with the rate expression:

\[ -\frac{1}{x} \frac{dA}{dt} = -\frac{1}{y} \frac{dB}{dt} = \frac{1}{z} \frac{dC}{dt} \]

### Step 1: Identify the stoichiometric coefficients
In the reaction, the stoichiometric coefficients are:
- For A: 1
- For B: 2
- For C: 3

### Step 2: Write the rate expressions
The rate of disappearance of reactants and the rate of formation of products can be expressed as follows:
- Rate of disappearance of A: \(-\frac{dA}{dt}\)
- Rate of disappearance of B: \(-\frac{1}{2} \frac{dB}{dt}\) (since the coefficient of B is 2)
- Rate of formation of C: \(\frac{1}{3} \frac{dC}{dt}\) (since the coefficient of C is 3)

### Step 3: Set up the rate expression
From the stoichiometric coefficients, we can express the rate of the reaction:
\[ -\frac{1}{1} \frac{dA}{dt} = -\frac{1}{2} \frac{dB}{dt} = \frac{1}{3} \frac{dC}{dt} \]

### Step 4: Compare with the given expression
Now, we can compare the coefficients from our derived expression with the given rate expression:
- From \(-\frac{1}{x} \frac{dA}{dt} = -\frac{1}{1} \frac{dA}{dt}\), we find that \(x = 1\).
- From \(-\frac{1}{y} \frac{dB}{dt} = -\frac{1}{2} \frac{dB}{dt}\), we find that \(y = 2\).
- From \(\frac{1}{z} \frac{dC}{dt} = \frac{1}{3} \frac{dC}{dt}\), we find that \(z = 3\).

### Conclusion
Thus, the values of \(x\), \(y\), and \(z\) are:
- \(x = 1\)
- \(y = 2\)
- \(z = 3\)

### Final Answer
The values of \(x\), \(y\), and \(z\) are:
- \(x = 1\)
- \(y = 2\)
- \(z = 3\)

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