The rate constant for the decomposition of gaseous azomethane,
C2 H6 N2 (g) > N2 (g) + C2 H6 (g)
Is 40.8 min-1 at 425OC. Find the number of moles azomethane and the number of moles of nitrogen 0.0500 minutes after 2.00 g of azomethane is introduced. Assume that the decomposition reaction is first order and that the reverse reaction can be ignored (Note that the first order integrated rate equation can be expressed in terms of the number of moles of the reacting species instead of the concentration).
Do we use the integrated law here?
Yes, in this case we can use the integrated law since the given rate constant corresponds to a first-order reaction. The integrated rate law for a first-order reaction is given by:
ln[A] = -kt + ln[A]₀
Where:
- [A] is the number of moles of the reacting species at time t
- k is the rate constant
- t is the time elapsed
- [A]₀ is the initial number of moles of the reacting species
To find the number of moles of azomethane and nitrogen 0.0500 minutes after 2.00 g of azomethane is introduced, we can follow these steps:
1. Convert the given mass of azomethane to moles. The molar mass of azomethane (C2 H6 N2) can be calculated by adding the atomic masses of carbon (C), hydrogen (H), and nitrogen (N). Assume the atomic masses of carbon, hydrogen, and nitrogen are 12.01 g/mol, 1.01 g/mol, and 14.01 g/mol, respectively.
- Calculate the molar mass: (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 14.01 g/mol)
- Convert the mass to moles: 2.00 g / molar mass
2. Use the integrated rate law to calculate the number of moles of azomethane at 0.0500 minutes.
- Plug in the values: ln[A] = -kt + ln[A]₀
[A] = e^(-kt + ln[A]₀)
- Substitute the given values:
- k = 40.8 min⁻¹
- t = 0.0500 minutes
- [A]₀ = initial moles of azomethane (calculated in step 1)
- Calculate [A] using the formula: [A] = e^(-k * t) * [A]₀
3. Calculate the number of moles of nitrogen produced using the stoichiometry of the reaction.
- The reaction stoichiometry tells us that for 1 mole of azomethane decomposed, 1 mole of nitrogen is produced. Therefore, the number of moles of nitrogen is equal to the number of moles of azomethane at 0.0500 minutes.
By following these steps, you can use the integrated law to find the number of moles of azomethane and nitrogen 0.0500 minutes after 2.00 g of azomethane is introduced.