The differential rate law for the reaction

Options

A. `(-d[H_2])/(dt) = (-d[I_2])/(dt) = (-d[HI])/(dt)`
B. `(d[H_2])/(dt) = (d[I_(2)])/(dt) = (d[HI])/(dt)`
C. `1/2 (d[H_(2)])/(dt) = (d[I_(2)])/(dt) = (-d[HI])/(dt)`
D. `-2 (d[H_(2)])/(dt) = -2 (d[I_(2)])/(dt) = (d[HI])/(dt)`

Explanation

To derive the differential rate law for the reaction \( H_2 + I_2 \rightarrow 2HI \), we will follow these steps:

### Step 1: Write the general form of the rate law
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products over time. For a general reaction:
\[ aA + bB \rightarrow cC + dD \]
The rate can be expressed as:
\[ \text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \]

### Step 2: Apply it to the given reaction
For the reaction \( H_2 + I_2 \rightarrow 2HI \), we identify:
- \( A = H_2 \) with \( a = 1 \)
- \( B = I_2 \) with \( b = 1 \)
- \( C = HI \) with \( c = 2 \)

### Step 3: Write the rate expressions for the reactants and products
Using the stoichiometric coefficients, we can write:
\[ \text{Rate} = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

### Step 4: Rearranging the equations
To express the rate in terms of the reactants, we multiply the rate expressions for \( H_2 \) and \( I_2 \) by their stoichiometric coefficients:
\[ -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

### Step 5: Combine the expressions
From the above, we can express the rate law as:
\[ -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

### Step 6: Finalize the rate law
To express the rate law in a more standard form, we can write:
\[ -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]

### Conclusion
Thus, the final differential rate law for the reaction \( H_2 + I_2 \rightarrow 2HI \) is:
\[ -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt} \]