I am having problems with these two questions.
1.The value of the equilibrium constant (Kp) as represented by the first chemical equation is 1.58 x 102 at 727 °C. Calculate the value of the equilibrium constant (Kp) for the second equation at the same temperature. Express answer in scientific notation.
2NO(g)+O2(g) = 2NO2(g)
2NO2(g) = 2NO(g)+O2(g)
2.When a sample of Cl(g) (16.72 mol) is placed in 340.0 L reaction vessel at 875.0 K and allowed to come to equilibrium the mixture contains 358.2 grams of Cl2(g). What concentration (mol/L) of Cl(g) reacted?
2Cl(g) = Cl2(g)
Unit Conversions:
K = C + 273
Molar Mass (g/mol)
Cl(g) 35.453
Cl2(g) 70.906
For #1, equation 2 is just the reverse of equation 1. So use K, which is given for equation 1, and take the reciprocal. K for equation 2 will be 1/K(eqn 1).
For #2, I would convert 358.2 g Cl2 to mols. Convert mols Cl2 to mols Cl, and that is mols Cl that reacted. That divided by 340.0 L = molarity = mols/L.
Check my thinking.
To solve these problems, we need to use the concept of equilibrium constant and the given data.
1. In the first question, we are given the equilibrium constant (Kp) for the first chemical equation at a specific temperature. We need to calculate the equilibrium constant (Kp) for the second equation at the same temperature.
The equilibrium constant (Kp) is determined by the ratio of the product concentrations to the reactant concentrations, with each concentration raised to the power of its stoichiometric coefficient. In this case, the stoichiometric coefficients are 1 for NO(g), 1 for O2(g), and 2 for NO2(g).
The first equation is:
2NO(g) + O2(g) ⇌ 2NO2(g)
The equilibrium constant (Kp1) for this equation is given as 1.58 x 10^2.
To calculate the equilibrium constant (Kp2) for the second equation, we can use the mathematical relationship:
Kp2 = (Kp1)^(1/2)
Here, we raise Kp1 to the power of 1/2 because the second equation is the reverse reaction of the first equation, and we reverse the stoichiometric coefficient.
So, Kp2 = (1.58 x 10^2)^(1/2)
Calculating this expression, the value of Kp2 is approximately 12.56 in scientific notation.
2. In the second question, we are given the amount of Cl(g) and the volume of the reaction vessel, and we need to calculate the concentration (mol/L) of Cl(g) that reacted.
First, we need to convert the mol of Cl(g) to the concentration in mol/L by dividing the mol of Cl(g) by the volume of the reaction vessel.
Given:
Mol of Cl(g) = 16.72 mol
Volume of reaction vessel = 340.0 L
Concentration of Cl(g) = Mol of Cl(g) / Volume of reaction vessel
Concentration of Cl(g) = 16.72 mol / 340.0 L
Calculating this expression, the concentration of Cl(g) is approximately 0.0491 mol/L.