c. A student put 1.18 mole of substance A and and 2.85 mole of substance B into a 10 litre flask which was then closed. The reaction that took place was: A(g)+ 2B(g) 3C(g) + D(g)
On analysis the equilibrium mixture at 25 0 C was found to contain 0.376mole of D. Calculate Kc and Kp at this temperature.
[ R = 0.0821atm l mole-1K-1].
These are in 1 L; therefore, mols = M.
........A + 2B ==> 3C + D
I .....1.18..2.85...0....0
C......-x....-2x...3x....x
E...1.18-x..2.85-x..3x...x
x in the problem = 0.376 for D at equilibrium.
Calculate concns of each of the others and substitute into Keq expression and solve for Kc.
Then convert top Kp = Kc*RTdelta n
I need help
yawa pota
To calculate Kc and Kp, we need to first determine the equilibrium concentrations of substances A, B, C, and D.
Using the stoichiometry of the reaction, we can determine that at equilibrium, the molar concentration of A will be 1.18 - x (where x is the molar concentration of A reacted), the molar concentration of B will be 2.85 - 2x (where 2x is the molar concentration of B reacted), the molar concentration of C will be 3x, and the molar concentration of D will be 0.376 + x.
Since we have the total volume and the number of moles, we can convert the concentrations to molarities (M). Molarity is defined as moles of a substance divided by the volume in liters (mol/L).
Given that the volume is 10 liters, we can divide the number of moles by the volume to find the molarities:
Molarity of A = (1.18 - x) / 10 M
Molarity of B = (2.85 - 2x) / 10 M
Molarity of C = (3x) / 10 M
Molarity of D = (0.376 + x) / 10 M
To calculate Kc, we use the equilibrium expression:
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b),
where [A], [B], [C], and [D] are the molar concentrations of A, B, C, and D, respectively.
In this case, since the stoichiometric coefficients for A, B, C, and D are all 1, the exponent values (a, b, c, d) are all 1.
Plugging in the molarities, we get:
Kc = ([3x/10] * [(0.376 + x)/10]) / ([(1.18 - x)/10] * [(2.85 - 2x)/10])
Now, let's simplify this equation:
Kc = (3x * (0.376 + x)) / ((1.18 - x) * (2.85 - 2x))
Next, we need to express Kp in terms of partial pressures.
Kp is related to Kc by the following equation:
Kp = Kc * (RT)^(∆n),
where R is the ideal gas constant, T is the temperature in Kelvin, and ∆n is the change in moles of gas during the reaction.
In this case, ∆n = (moles of product gases) - (moles of reactant gases) = (1 + 1) - (1 + 2) = -1.
Plugging in the values, we get:
Kp = Kc * (R * T)^(-1)
= Kc / (R * T)
Given that R = 0.0821 atm L/mol K and T = 25°C = 298K, we can calculate Kp as follows:
Kp = Kc / (0.0821 * 298)
By inserting the calculated values of Kc and Kp, we can obtain the final solutions for each.