For the reaction : `H_(2)+I_(2)to 2HI,` the differential rate law is

Options

A. `-(d[H_2])/(dt)=-(d[l_2])/(dt)=1/2(d[Hl])/(dt)`
B. `-(d[H_2])/(dt)=-2(d[l_2])/(dt)=1/2(d[Hl])/(dt)`
C. `-(d[H_2])/(dt)=-(d[l_2])/(dt)=(d[Hl])/(dt)`
D. `-(d[H_2])/(dt)=-(d[l_2])/(dt)=-(d[Hl])/(dt)`

Answer

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

### Step 1: Understand the Reaction
The reaction involves hydrogen gas (\( H_2 \)) and iodine gas (\( I_2 \)) combining to form hydrogen iodide (\( HI \)).

### Step 2: Write the Rate of Reaction
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products over time. For this reaction, we can express the rate in terms of the reactants and the product:

\[ \text{Rate} = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2}\frac{d[HI]}{dt} \]

### Step 3: Establish the Differential Rate Law
From the above expression, we can see that the rate of the reaction is related to the change in concentration of the reactants and products. Since the stoichiometry of the reaction is 1:1:2, we can express the rate law as:

\[ \text{Rate} = k[H_2]^m[I_2]^n \]

Where \( k \) is the rate constant, and \( m \) and \( n \) are the orders of the reaction with respect to \( H_2 \) and \( I_2 \), respectively.

### Step 4: Determine the Orders of the Reaction
For this specific reaction, if we assume it is elementary (which means the rate law can be directly derived from the stoichiometry), we can say:

- The order with respect to \( H_2 \) is 1 (since it appears once in the balanced equation).
- The order with respect to \( I_2 \) is also 1.

Thus, the differential rate law can be expressed as:

\[ \text{Rate} = k[H_2]^1[I_2]^1 = k[H_2][I_2] \]

### Final Expression
Therefore, the differential rate law for the reaction \( H_2 + I_2 \rightarrow 2HI \) is:

\[ \text{Rate} = k[H_2][I_2] \]