For the reaction below, Kc = 0.154 at 2000 K.
2 CH4(g) ⇌ C2H2(g) + 3 H2(g)
A 0.750 L equilibrium mixture at 2000 K contains 0.650 mol each of CH4(g) and H2(g).
What is the equilibrium concentration of C2H2(g)?
Well, it seems like we have a little chemical party going on here! So, let's break it down.
First, let's start with what we know. We have an equilibrium mixture with 0.650 mol of CH4(g) and H2(g) each. Since our reaction is 2 CH4(g) ⇌ C2H2(g) + 3 H2(g), we know that the stoichiometry tells us that for every 2 moles of CH4, we get 1 mole of C2H2.
In our case, we have 0.650 mol of CH4, which means we will form half of that amount in moles of C2H2. Therefore, we will have 0.325 mol of C2H2.
Now that we have the moles of C2H2, we can calculate the equilibrium concentration. Since we have a 0.750 L equilibrium mixture, we divide the moles of C2H2 by the volume to get the molarity.
Concentration of C2H2 = (0.325 mol) / (0.750 L) = 0.433 M
So, the equilibrium concentration of C2H2(g) in the mixture is 0.433 M.
To determine the equilibrium concentration of C2H2(g), we need to use the equilibrium constant expression and the known concentrations of CH4(g) and H2(g).
The equilibrium constant expression for the given reaction is:
Kc = [C2H2(g)] / ([CH4(g)]^2 * [H2(g)]^3)
Given:
- Kc = 0.154
- [CH4(g)] = 0.650 mol / 0.750 L = 0.867 M
- [H2(g)] = 0.650 mol / 0.750 L = 0.867 M
We need to find [C2H2(g)].
Let's substitute the given values into the equilibrium constant expression:
0.154 = [C2H2(g)] / [(0.867 M)^2 * (0.867 M)^3]
Simplifying the denominator:
(0.867 M)^2 * (0.867 M)^3 = 0.867^5 M^5 = 0.489 M^5
Now, rearrange the equation and solve for [C2H2(g)]:
[C2H2(g)] = 0.154 * 0.489 M^5
[C2H2(g)] ≈ 0.0753 M
Therefore, the equilibrium concentration of C2H2(g) is approximately 0.0753 M.
To find the equilibrium concentration of C2H2(g), we need to use the equilibrium constant expression and the given information.
First, let's write down the balanced equation again:
2 CH4(g) ⇌ C2H2(g) + 3 H2(g)
The equilibrium constant expression, Kc, is given by:
Kc = [C2H2(g)] / ([CH4(g)]^2 * [H2(g)]^3)
In this case, we are given that Kc = 0.154 and the initial concentrations of CH4(g) and H2(g).
Let's use the information provided to set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations at equilibrium.
Initial concentrations:
[CH4(g)] = 0.650 mol
[H2(g)] = 0.650 mol
[C2H2(g)] = 0 (since it's not given)
Change in concentrations:
Let's use the variable "x" to represent the change in concentration of C2H2(g). Since two moles of CH4(g) produce one mole of C2H2(g), the change in [C2H2(g)] will be equal to 2x. Similarly, the change in [H2(g)] will be equal to 3x.
Equilibrium concentrations:
[CH4(g)] = 0.650 mol - x
[H2(g)] = 0.650 mol - 3x
[C2H2(g)] = 2x
Now, let's substitute the equilibrium concentrations into the equilibrium constant expression:
Kc = [C2H2(g)] / ([CH4(g)]^2 * [H2(g)]^3)
0.154 = (2x) / ((0.650 - x)^2 * (0.650 - 3x)^3)
Now, we need to solve this equation to find the value of x, and then substitute it back to find the equilibrium concentration of C2H2(g).
This equation can be solved by rearranging, simplifying, and then using numerical techniques such as a calculator or a computer program. Unfortunately, the steps involved in solving the equation are beyond my capabilities as a text-based AI. However, you can utilize a calculator, algebraic software, or numerical techniques to solve for x and determine the equilibrium concentration of C2H2(g) (2x) based on the equation above.
A 0.750 L equilibrium mixture at 2000 K contains 0.650 mol each of CH4(g) and H2(g).
(CH4) = (C2H2) = 0.650 mols/0.750 L = approx 0.9 but you need to use a more accurate number.
....................2 CH4(g) ⇌ C2H2(g) + 3 H2(g)
E....................0.9....................x..............0.9
Kc = (C2H2)(H2)^3/(CH4)^2
You know Kc, H2, and CH4, solve for C2H2