A substance (CD) decomposes into C and D
At the temperature of the experiment, 15.0% of CD is decomposed when equilibrium is established.
a) if the initial concentration of CD is 0.200mol/L, what are the equilibrium concentrations of CD, C and D?
I got 0.17 for CD and I got 0.015 for C and D.
b) What is K for the reaction at this temperature?
I need help with b please :)
Thank you for posting your work. You also need help with part a.
If CD ==> C + D, then for every 1 mol CD that decomposes, you obtain 1 mol D and 1 mol C; therefore, 0.2 x 0.15 = 0.03 and the (C) = (D) = 0.03. 0.17 is correct for CD.
2.
Keq = (C)(D)/(CD) = (0.03)(o.03)/(0.17) = ?
Oh now it makes sense. Thank you so much DrBob222 :)
To determine the equilibrium constant (K) for the reaction at this temperature, we need to set up and solve the expression for K based on the given concentrations at equilibrium.
The balanced equation for the reaction is:
CD ⇌ C + D
Let's denote the equilibrium concentrations of CD, C, and D as [CD], [C], and [D], respectively.
According to the information given, 15.0% of CD is decomposed at equilibrium. This means that 85.0% of CD remains unchanged. Therefore, the equilibrium concentration of CD can be calculated as:
[CD] = (initial concentration of CD) - (amount decomposed)
= 0.200 mol/L - (0.150 * 0.200 mol/L)
= 0.200 mol/L - 0.030 mol/L
= 0.17 mol/L
The equilibrium concentration of C and D can be determined using stoichiometry. Since the stoichiometry of the reaction is 1:1, the equilibrium concentrations of C and D would be equal to the amount decomposed, which is 0.015 mol/L.
Therefore, the equilibrium concentrations are:
[CD] = 0.17 mol/L
[C] = 0.015 mol/L
[D] = 0.015 mol/L
Now let's move on to calculating the equilibrium constant (K).
The equilibrium constant expression (Kc) for the reaction can be written as:
Kc = [C] * [D] / [CD]
Substituting the values we have obtained:
Kc = (0.015 mol/L) * (0.015 mol/L) / (0.17 mol/L)
= 0.000225 mol^2/L^2 / 0.17 mol/L
= 0.00132352941 L/mol
At this temperature, the equilibrium constant (Kc) for the reaction is approximately 0.0013 L/mol.
Note: The equilibrium constant (Kc) is unitless because the concentrations are divided by the concentration of CD, and it cancels out in the expression.
To determine the value of K for the reaction, we need to use the equilibrium concentrations of the reactants and products.
The given information states that when equilibrium is established, 15.0% of CD is decomposed. This means that 85.0% of CD remains in equilibrium.
Since the initial concentration of CD is 0.200 mol/L, we can calculate the equilibrium concentration of CD as follows:
CD (equilibrium) = 85.0% of 0.200 mol/L = 0.85 * 0.200 mol/L = 0.170 mol/L
Now, we need to determine the equilibrium concentration of C and D. According to the reaction, for every one mole of CD that decomposes, one mole of C is formed and one mole of D is formed. Since 15.0% of CD decomposes, the equilibrium concentrations of C and D are also 0.015 mol/L.
Finally, we can determine the value of K using the equilibrium concentrations of C, D, and CD.
K = [C]^c * [D]^d / [CD]^cd
In this case, since the stoichiometric coefficient of C, D, and CD is 1, we have:
K = [C][D] / [CD]
Substituting the equilibrium concentrations we found earlier, we get:
K = (0.015 mol/L)(0.015 mol/L) / (0.170 mol/L)
Calculating this expression, we find:
K = 0.00135 mol^2/L^2
Therefore, the value of K for the reaction at this temperature is 0.00135 mol^2/L^2.