The half-life of first order decomposition of `NH_(4)NO_(3)` is 2.10 hr at 288 K temperature
`NH_(4)NO_(3)(aq)toN_(2)O(g)+2H_(2)O(l)`. If 6.2 of `NH_(4)NO_(3)` is allowed to decompose, the required for `NH_(4)NO_(3)` to decompose 90 % is :

Options

A. 6.978 hr
B. 0.319
C. 0.319 hr
D. None of these

Answer

To solve the problem of determining the time required for 90% decomposition of `NH4NO3`, we can follow these steps:

### Step 1: Calculate the Rate Constant (k)

The half-life (\(t_{1/2}\)) of a first-order reaction is given by the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

Given that the half-life of the decomposition of `NH4NO3` is 2.10 hours, we can rearrange the formula to find \(k\):

\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{2.10 \text{ hours}} \]

Calculating this gives:

\[ k \approx 0.329 \text{ hr}^{-1} \]

### Step 2: Set Up the First-Order Kinetics Equation

For a first-order reaction, the relationship between the initial concentration (\(A_0\)), the concentration at time \(t\) (\(A_t\)), and the rate constant \(k\) is given by:

\[ \ln\left(\frac{A_0}{A_t}\right) = kt \]

### Step 3: Determine Initial and Final Concentrations

In this case, we can assume the initial concentration of `NH4NO3` is 100% (or we can denote it as 100 units). After 90% decomposition, only 10% remains. Thus:

- \(A_0 = 100\)
- \(A_t = 10\)

### Step 4: Substitute Values into the Equation

Substituting the values into the first-order kinetics equation:

\[ \ln\left(\frac{100}{10}\right) = kt \]

This simplifies to:

\[ \ln(10) = kt \]

### Step 5: Solve for Time (t)

Now we can substitute the value of \(k\) we calculated earlier:

\[ \ln(10) \approx 2.303 \]

Thus, we have:

\[ 2.303 = (0.329 \text{ hr}^{-1}) \cdot t \]

Now, solving for \(t\):

\[ t = \frac{2.303}{0.329} \approx 6.978 \text{ hours} \]

### Final Answer

The time required for `NH4NO3` to decompose 90% is approximately **6.978 hours**.

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