carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is S2(g)+C(S)=CS2(g) and Kc = 9.4 at 900K.
How many grams of CS2(g) can be prepared by heating 8.9moles of S2(g) with excess carbon in a 5.00L reaction vessel held at 900K until equilibrium is attained?
(S2) = mols/L = 8.9/5.00 = approx 1.8 but you need a better number than that. Recalculate all that follows.
......S2 + C(s) ==> CS2
I....1.8............0
C.....-x............x
E....1.8-x..........x
Note that C does not appear in the Kc expression because it is a solid.
Kc = (CS2)/(S2) = 9.4
Substitute and solve for x. I get approx 2, then mols = M x L = 5*2 = about 10 and grams = mols x molar mass CS2. Remember all the number I have placed above are estimates. Post your work if you get stuck.
To determine the grams of CS2(g) that can be prepared, we first need to understand the given chemical equation and equilibrium constant (Kc) value.
The balanced chemical equation for the reaction is:
S2(g) + C(s) → CS2(g)
The equilibrium constant expression (Kc) can be written as:
Kc = [CS2(g)] / ([S2(g)] * [C(s)])
Given Kc = 9.4, we can set up the equilibrium expression:
9.4 = [CS2(g)] / ([S2(g)] * [C(s)])
Since the reaction vessel is held at 900K until equilibrium is attained, we know that the initial moles of S2(g) will be 8.9 moles. Additionally, carbon is provided in excess, indicating that it will not be a limiting reactant.
Now, let's calculate the concentration of S2(g) and C(s) in terms of moles per liter (M).
Concentration of S2(g) = moles of S2(g) / volume of reaction vessel
Concentration of S2(g) = 8.9 moles / 5.00 L
Concentration of S2(g) = 1.78 M
Concentration of C(s) = since carbon is in excess, its concentration will not change throughout the reaction.
Using the equilibrium expression, we can rearrange it to solve for [CS2(g)]:
9.4 = [CS2(g)] / (1.78 M * [C(s)])
Since [C(s)] is constant, we can represent it as a single constant term "k":
9.4 = [CS2(g)] / (1.78 M * k)
To find [CS2(g)], we rearrange the equation:
[CS2(g)] = 9.4 * (1.78 M * k)
Now, substitute the value of [CS2(g)] back into the original equation to determine the grams of CS2(g):
grams of CS2(g) = [CS2(g)] * molar mass of CS2(g)
The molar mass of CS2(g) is the sum of the atomic masses of carbon and sulfur:
Molar mass of CS2(g) = (12.01 g/mol) + (32.07 g/mol) + (32.07 g/mol)
Molar mass of CS2(g) = 76.15 g/mol
Finally, calculate the grams of CS2(g):
grams of CS2(g) = [CS2(g)] * 76.15 g/mol