A chemist is studying the following equilibrium:
2Pb(NO3)2 (s) ! 2PbO (s) + 4NO2 (g) + O2 (g) He starts out with 10 g of Pb(NO3)2 and, at equilibrium, has 2.02 g of PbO. The concentrations of NO2 and O2 at equilibrium are 0.25 M and 0.019 M, respectively. What is the value of the equilibrium constant?
Im in the same boat lmao
To find the value of the equilibrium constant for the given equilibrium, we need to use the given information about the initial and equilibrium amounts of the substances involved.
Step 1: Calculate the moles of each substance.
- The molar mass of Pb(NO3)2 is 331.2 g/mol, so 10 g of Pb(NO3)2 is equal to 10 g / 331.2 g/mol = 0.0302 mol.
- The molar mass of PbO is 223.2 g/mol, so 2.02 g of PbO is equal to 2.02 g / 223.2 g/mol = 0.00907 mol.
Step 2: Set up the equilibrium expression.
The equilibrium expression for the given reaction is:
Kc = ([PbO]^2 * [NO2]^4 * [O2]) / ([Pb(NO3)2]^2)
Where [PbO], [NO2], and [O2] represent the equilibrium concentrations of PbO, NO2, and O2 respectively, and [Pb(NO3)2] represents the initial concentration of Pb(NO3)2.
Step 3: Plug in the values into the equilibrium expression.
Kc = (0.00907^2 * (0.25)^4 * (0.019)) / (0.0302^2)
Kc = 0.000294
Therefore, the value of the equilibrium constant (Kc) is approximately 0.000294.