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_(2) = k_(1) = k_(3) = k_(4)`
B. `k_(2) = 2k_(1) = k_(3) = k_(4)`
C. `k_(2) = 2k_(3) = k_(1) = k_(4)`
D. `k_(2) = k_(1) = 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 can analyze the rate expressions given in the problem.

### 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: Analyze the Stoichiometry of the Reaction

The stoichiometry of the reaction shows that for every 2 moles of \( NO \) consumed, 2 moles of \( H_2 \) are consumed, and 1 mole of \( N_2 \) and 2 moles of \( H_2O \) are produced.

### Step 3: Relate the Rate Expressions

From the stoichiometry, we can relate the rates of change of the concentrations:

- The rate of formation of \( N_2 \) is half the rate of consumption of \( NO \) because 2 moles of \( NO \) produce 1 mole of \( N_2 \):

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

- Similarly, the rate of formation of \( H_2O \) is equal to the rate of consumption of \( NO \):

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

### Step 4: Substitute the Rate Expressions

Now, substituting the rate expressions into the relationships we derived:

1. From \( \frac{d[N_2]}{dt} = k_1[NO][H_2] \) and \( -\frac{d[NO]}{dt} = k_3[NO][H_2] \):
\[ k_1[NO][H_2] = \frac{1}{2} k_3[NO][H_2] \]
Therefore, we can write:
\[ k_3 = 2k_1 \]

2. From \( \frac{d[H_2O]}{dt} = k_2[NO][H_2] \) and \( -\frac{d[NO]}{dt} = k_3[NO][H_2] \):
\[ k_2[NO][H_2] = -\frac{d[NO]}{dt} = k_3[NO][H_2] \]
Thus, we have:
\[ k_2 = k_3 \]

3. From \( -\frac{d[H_2]}{dt} = k_4[NO][H_2] \) and \( -\frac{d[NO]}{dt} = k_3[NO][H_2] \):
\[ k_4[NO][H_2] = k_3[NO][H_2] \]
Therefore, we can conclude:
\[ k_4 = k_3 \]

### Step 5: Final Relationships

From the above relationships, we summarize:

- \( k_3 = 2k_1 \)
- \( k_2 = k_3 \)
- \( k_4 = k_3 \)

Thus, we can express all rate constants in terms of \( k_1 \):
- \( k_2 = k_3 = k_4 = 2k_1 \)

### Conclusion

The relationship between the rate constants is:

\[ k_2 = k_3 = k_4 = 2k_1 \]