Asked by Mark
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
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
Answers
Answered by
DrBob222
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).
Answered by
DrBob222
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.
Check my thinking.