Question

The half-life of first order decomposition of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(2.10 \mathrm{hr}\) at \(288 \mathrm{~K}\) temperature \(\mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\), if \(6.2 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is allowed to decompose, The time required for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) to decompose \(90 \%\) and the volume of dry \(\mathrm{N}_{2} \mathrm{O}\) produced at this point measured at STP are respectively: (a) \(6.978 \mathrm{hr}, 2.016 \mathrm{~L}\) (b) \(0.319 \mathrm{hr}, 2.12 \mathrm{~L}\) (c) \(0.319 \mathrm{hr}, 2.016 \mathrm{~L}\) (d) None of these

Short answer

\(6.978 \) hours, \(2.016 \) L

Step-by-step solution

Determine time required for 90% decomposition

Use the half-life formula for first order reactions: \( t = \frac{\ln(\frac{[A]_0}{[A]})}{k} \), where \( t \) is the time required, \( [A]_0 \) is the initial amount, \( [A] \) is the final amount, and \( k \) is the rate constant. Since 90% is decomposed, only 10% remains: \( [A] = 0.10[A]_0 \). We need to find \( k \) using half-life formula for first order reactions: \( t_{1/2} = \frac{\ln(2)}{k} \), then solve for \( t \).

Calculate the rate constant, k

From the half-life equation \( t_{1/2} = \frac{\ln(2)}{k} \), solve for \( k \): \( k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{2.10} \) hours\(^{-1}\).

Calculate the time for 90% decomposition

Now using the equation from Step 1 and the value of \( k \) from Step 2, calculate \( t \): \( t = \frac{\ln(\frac{[A]_0}{0.10[A]_0})}{k} = \frac{\ln(10)}{k} \). Substitute the value of \( k \) to find the time required.

Calculating the amount of \(N_2O\) produced

For every mole of \(NH_4NO_3\) decomposed, one mole of \(N_2O\) is produced. Calculate the initial moles of \(NH_4NO_3\) using its molar mass, and find the moles of \(N_2O\) produced after 90% decomposition.

Calculate the volume of \(N_2O\) at STP

Using the ideal gas law at STP conditions (1 mole of gas occupies 22.4 L), calculate the volume of the \(N_2O\) produced: Volume = moles of \(N_2O\) \(\times\) 22.4 L/mol.

Key concepts

Chemical Kinetics and First-Order Reactions

Chemical kinetics is the area of chemistry concerned with understanding the rates of chemical reactions. It is crucial for the prediction of the behavior of chemical systems and for the design of chemical processes. A deep-dive into this field reveals that reactions can be classified based on their order, and a first-order reaction depends linearly on the concentration of a single reactant.

For first-order reactions, the half-life, which is the time it takes for half of the reactant to be converted into products, remains constant regardless of the initial concentration. This is an important feature and is mathematically expressed as the half-life formula for first order reactions: \( t_{1/2} = \frac{\ln(2)}{k} \) where \( k \) is the rate constant of the reaction. Understanding this concept helps in solving problems where reaction time and extent are critical,—for example, in pharmaceuticals, where drug clearance from the body often follows first-order kinetics.

Rate Constant Calculation for First-Order Reactions

The rate constant, \( k \), is a crucial parameter in chemical kinetics that determines the speed of a reaction. For a first-order reaction, it can be determined if the half-life of the reaction is known. The calculation uses the relation \( k = \frac{\ln(2)}{t_{1/2}} \), which derives from the first-order reaction kinetics principles.

To extract the most information from this equation, it's important to note that \( \ln(2) \) is a constant arising from the natural logarithm of 2, as half-life deals with a 50% conversion rate. The rate constant is inversely proportional to the half-life: a shorter half-life translates to a larger rate constant, implying a faster reaction. This becomes particularly convenient when comparing different reactions or conditions to understand which proceed more rapidly.

Applying the Ideal Gas Law at STP

STP, or standard temperature and pressure, is a set of conditions often used as a reference to predict gas behaviors. It's defined as 0°C (273.15 K) and 1 atmosphere pressure. The ideal gas law, given by \( PV = nRT \), is a fundamental equation that relates the variables of pressure (\( P \)), volume (\( V \)), moles of gas (\( n \)), gas constant (\( R \) — 0.0821 L·atm/mol·K), and temperature (\( T \) in Kelvin).

At STP, the law simplifies the calculations because each mole of any ideal gas occupies 22.4 liters. For example, to find the volume of a gas produced in a chemical reaction at STP, you would first calculate the number of moles of gas produced and then multiply by 22.4 L/mol. This relationship is vital for laboratory and industrial applications where the production of gases under standard conditions is a common requirement.