For the reaction,
`2NO(g) + 2H_(2)(g) rarr N_(2)(g) + 2H_(2)O(g)`
The rate expression can be written in the following ways:
`(d[N_(2)])/(d t) = k_(1)[NO][H_(2)], (d[H_(2)O])/(d t) = k_(2)[NO][H_(2)]`
`- (d[NO])/(d t) = k_(3)[NO][H_(2)], -(d[H_(2)])/(d t) = k_(4)[NO] [H_(2)]`
The relationship between `k_(1), k_(2), k_(3), k_(4)` is

Options

A. `k_(1)=k_(2)=k_(3)=k_(4)`
B. `k_(1)=2k_(2)=k_(3)=k_(4)`
C. `2k_(1)=k_(2)=k_(3)=k_(4)`
D. `2k_(1)=2k_(2)=k_(3)=2k_(4)`

Explanation

To find the relationship between the rate constants \( k_1, k_2, k_3, \) and \( k_4 \) for the reaction

\[ 2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g), \]

we will analyze the given rate expressions step by step.

### Step 1: Write the rate expressions
The rate expressions provided are:

1. \(\frac{d[N_2]}{dt} = k_1 [NO][H_2]\)
2. \(\frac{d[H_2O]}{dt} = k_2 [NO][H_2]\)
3. \(-\frac{d[NO]}{dt} = k_3 [NO][H_2]\)
4. \(-\frac{d[H_2]}{dt} = k_4 [NO][H_2]\)

### Step 2: Relate the rates of formation and consumption
From stoichiometry, we know that the rates of formation and consumption are related by the coefficients in the balanced equation. Therefore, we can express the rates in terms of a common rate \( R \):

\[ R = \frac{1}{2} \left(-\frac{d[NO]}{dt}\right) = \frac{1}{2} \left(-\frac{d[H_2]}{dt}\right) = \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt} \]

### Step 3: Establish relationships between \( k_1, k_2, k_3, \) and \( k_4 \)
1. From the first and third expressions:
\[ R = \frac{1}{2} k_3 [NO][H_2] \quad \text{and} \quad R = k_1 [NO][H_2] \]
Equating these gives:
\[ k_1 [NO][H_2] = \frac{1}{2} k_3 [NO][H_2] \]
Canceling \([NO][H_2]\) (assuming they are not zero):
\[ k_3 = 2k_1 \]

2. From the second and fourth expressions:
\[ R = \frac{1}{2} k_4 [NO][H_2] \quad \text{and} \quad R = \frac{1}{2} k_2 [NO][H_2] \]
Equating these gives:
\[ \frac{1}{2} k_4 [NO][H_2] = \frac{1}{2} k_2 [NO][H_2] \]
Canceling \([NO][H_2]\):
\[ k_4 = k_2 \]

3. From the first and second expressions:
\[ R = k_1 [NO][H_2] \quad \text{and} \quad R = \frac{1}{2} k_2 [NO][H_2] \]
Equating these gives:
\[ k_1 [NO][H_2] = \frac{1}{2} k_2 [NO][H_2] \]
Canceling \([NO][H_2]\):
\[ k_2 = 2k_1 \]

### Step 4: Summarize the relationships
From the above steps, we have established the following relationships:
- \( k_3 = 2k_1 \)
- \( k_4 = k_2 \)
- \( k_2 = 2k_1 \)

Thus, we can conclude:
\[ k_3 = k_4 = 2k_1 \]

### Final Relationship
Putting it all together, we find:
\[ k_2 = 2k_1, \quad k_3 = 2k_1, \quad k_4 = 2k_1 \]
So, the final relationship can be summarized as:
\[ k_2 = k_3 = k_4 = 2k_1 \]