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. `-0.2 xx 10^(-4) Ms^(-1)`
B. `0.2 xx 10^(-4) Ms^(-1)`
C. `0.1 xx 10^(-4) Ms^(-1)`
D. `0.3 xx 10^(4) Ms^(-1)`

Answer

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

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

given that the rate of change of concentration of hydrogen (H₂) is \( 0.3 \times 10^{-4} \, \text{Ms}^{-1} \), we can follow these steps:

### Step 1: Identify the stoichiometric coefficients
From the balanced chemical equation, we can see the stoichiometric coefficients:
- For \( H_2 \), the coefficient is 3.
- For \( NH_3 \), the coefficient is 2.

### Step 2: Write the rate expressions
The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products. For the reaction, we have:

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

### Step 3: Substitute the given rate of change of hydrogen
Given that:

\[ \frac{d[H_2]}{dt} = -0.3 \times 10^{-4} \, \text{Ms}^{-1} \]

we can substitute this into the rate expression:

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

### Step 4: Calculate the rate of change of ammonia
Now, simplifying the left side:

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

Calculating the left side:

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

So we have:

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

### Step 5: Solve for \(\frac{d[NH_3]}{dt}\)
To find \(\frac{d[NH_3]}{dt}\), we multiply both sides by 2:

\[ \frac{d[NH_3]}{dt} = 2 \times 0.1 \times 10^{-4} = 0.2 \times 10^{-4} \, \text{Ms}^{-1} \]

### Final Answer
Thus, the rate of change of concentration of ammonia is:

\[ \frac{d[NH_3]}{dt} = 0.2 \times 10^{-4} \, \text{Ms}^{-1} \]

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