Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen 14) monoxide and oxygen according to the following chemical equation.
2 NO2(g) → 2 NO(g) + O2(g). A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and
the rate constant at 300°C are 11 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?
.42 other 2 are wrong didnt account for mole ratios
.84 mol
.084 mols
To solve this question, we can use the second-order reaction equation and the given half-life and rate constant to find the initial moles of NO2 in the sample.
We know that the half-life (t1/2) of a second-order reaction can be related to the rate constant (k) and the initial concentration of the reactant (A0) using the following equation:
t1/2 = 1 / (k * A0)
Given that the half-life is 11 seconds and the rate constant is 0.54 M-1 s-1, we can rearrange the equation to solve for A0:
A0 = 1 / (k * t1/2)
Substituting the given values:
A0 = 1 / (0.54 M-1 s-1 * 11 s)
A0 ≈ 0.1667 M
Now we need to convert the initial concentration to moles of NO2. To do this, we'll use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. In this case, we are given the volume (2.50 L) and the temperature (300°C, which we must convert to Kelvin).
To convert from Celsius to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
So, T(K) = 300°C + 273.15 = 573.15 K
Now we can rearrange the ideal gas law equation to solve for n:
n = PV / (RT)
Substituting the given values:
n = (1 atm * 2.50 L) / (0.0821 atm·L/mol·K * 573.15 K)
n ≈ 0.108 mol
Therefore, there were approximately 0.108 moles of NO2 in the original sample.