Oxides of nitrogen are produced in high-temperature combustion processes. The essential reaction is:
N2(g)+O2(g)--->2NO(g)
At what approximate temperature will an equimolar mixture of N2(g) and O2(g) be 5.0 % converted to NO(g)?
To find the approximate temperature at which an equimolar mixture of N2(g) and O2(g) is 5.0% converted to NO(g), we need to use the concept of equilibrium constant and the given reaction equation.
The equilibrium constant (K) of a reaction is defined as the ratio of the concentration of the products to the concentration of the reactants, each raised to their stoichiometric coefficients.
For the given reaction: N2(g) + O2(g) --> 2NO(g)
The equilibrium constant expression (K) can be written as:
K = [NO]^2 / [N2] [O2]
Here, [NO], [N2], and [O2] represent the concentrations of NO(g), N2(g), and O2(g) respectively.
Since the reaction mentioned in the question is an equimolar mixture, we can assume the initial concentration of N2(g) and O2(g) to be the same, let's say x. Thus, the concentration of NO(g) at equilibrium would be 2 * 0.05x = 0.1x.
Now, substituting these values into the equilibrium constant expression, we get:
K = (0.1x)^2 / x * x
Simplifying further:
K = 0.01x^2 / x^2
K = 0.01
We can now use the equilibrium constant expression to find the approximate temperature. Since temperature affects the equilibrium constant, we need the equilibrium constant expression that includes temperature, which is given by the Van't Hoff equation:
ln(K2 / K1) = -ΔH / R * (1 / T2 - 1 / T1)
Where:
K2 and K1 are the equilibrium constants at temperatures T2 and T1 respectively,
ΔH is the enthalpy change of the reaction,
R is the ideal gas constant (8.314 J/(mol·K)),
T2 and T1 are the absolute temperatures in Kelvin.
Now, let's assume the initial temperature (T1) to be known, and we want to find T2. Since we only need an approximate value, we can assume the heat of reaction (ΔH) to be constant. Let's say the value of ΔH is -200 kJ/mol.
Rearranging the Van't Hoff equation to solve for T2:
1 / T2 = 1 / T1 - (ln(K2 / K1) * R / ΔH)
Substituting the values:
1 / T2 = 1 / T1 - (ln(0.01 / K1) * R / -200000)
By rearranging the equation and solving for T2, you can approximate the temperature at which the equimolar mixture of N2(g) and O2(g) is 5.0% converted to NO(g).
Please note that this is just an approximate calculation, and it assumes several assumptions and simplifications. The actual temperature may vary depending on other factors and conditions.