i have difficulties in this question.any help would be appreciated.
1) 2No(g)-> N2(g)+O2(g)
at 1800K,K is 8.36*10^3.what concentration of NO(g) is in equilibrium with N2(g) at a concentration of 0.0500mol/liter and O2(g) at a concentration of 0.500 mol/liter?
The equilibrium constant expression is:
K(eq)=[N2][O2] / [NO]^2 = 8.36*10^3
Substitute the concentrations given and solve for [NO]
To determine the concentration of NO(g) in equilibrium with N2(g) and O2(g), we need to use the equilibrium constant expression for the given reaction. The equilibrium constant (K) is expressed as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.
In this case, the balanced equation is:
2NO(g) -> N2(g) + O2(g)
The equilibrium constant expression is:
K = [N2(g)] * [O2(g)] / [NO(g)]^2
Given concentrations:
[N2(g)] = 0.0500 mol/liter
[O2(g)] = 0.500 mol/liter
We need to find the concentration of [NO(g)].
To solve for [NO(g)], rearrange the equation:
[NO(g)]^2 = ([N2(g)] * [O2(g)]) / K
Substitute the given values:
[NO(g)]^2 = (0.0500 mol/liter * 0.500 mol/liter) / (8.36 * 10^3)
Calculate the numerator:
(0.0500 mol/liter * 0.500 mol/liter) = 0.025 mol^2/liter^2
Divide by K:
[NO(g)]^2 = 0.025 mol^2/liter^2 / (8.36 * 10^3)
Calculate the right-hand side of the equation:
0.025 mol^2/liter^2 / (8.36 * 10^3) ≈ 2.99 * 10^-6 mol^2/liter^2
Take the square root of both sides to solve for [NO(g)]:
[NO(g)] = √(2.99 * 10^-6 mol^2/liter^2)
[NO(g)] ≈ 0.00173 mol/liter
Therefore, the concentration of NO(g) in equilibrium with N2(g) at a concentration of 0.0500 mol/liter and O2(g) at a concentration of 0.500 mol/liter is approximately 0.00173 mol/liter.