For the reaction
`N_(2) + 3 H_(2) to 2 NH_(3)` The rate of change of concentration for hydrogen is `-0.3 xx 10^(-4) Ms^(-1)` The rate of change of concentration of ammonia is :

Options

A. `-0.2xx10^4`
B. `0.2xx10^(-4)`
C. `0.1xx10^(-4)`
D. `0.3xx10^(-4)`

Answer

To find the rate of change of concentration of ammonia (NH₃) for the reaction:

\[ N_2 + 3 H_2 \rightarrow 2 NH_3 \]

we start by analyzing the stoichiometry of the reaction.

### Step 1: Write the rate expressions based on stoichiometry.

From the balanced equation, we can express the rates of change of concentration for each species involved in the reaction. The rate of the reaction can be expressed as:

\[ -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

### Step 2: Substitute the known rate of change for hydrogen.

We are given that the rate of change of concentration for hydrogen is:

\[ \frac{d[H_2]}{dt} = -0.3 \times 10^{-4} \, \text{M/s} \]

Substituting this value into the rate expression gives:

\[ -\frac{1}{3} \left(-0.3 \times 10^{-4}\right) = \frac{1}{2} \frac{d[NH_3]}{dt} \]

### Step 3: Simplify the equation.

This simplifies to:

\[ \frac{0.3 \times 10^{-4}}{3} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

Calculating the left side:

\[ \frac{0.3 \times 10^{-4}}{3} = 0.1 \times 10^{-4} \]

So we have:

\[ 0.1 \times 10^{-4} = \frac{1}{2} \frac{d[NH_3]}{dt} \]

### Step 4: Solve for the rate of change of concentration of ammonia.

Now, multiply both sides by 2 to isolate \(\frac{d[NH_3]}{dt}\):

\[ \frac{d[NH_3]}{dt} = 2 \times 0.1 \times 10^{-4} = 0.2 \times 10^{-4} \, \text{M/s} \]

### Conclusion

Thus, the rate of change of concentration of ammonia is:

\[ \frac{d[NH_3]}{dt} = 0.2 \times 10^{-4} \, \text{M/s} \]