NOCL(g) decomposes to from nitrogen monoxide gas and chlorine gas. The equation is
2NOCL ----> 2NO +Cl2
At a certain temperature the equilibrium constant is 1.60 x 10^-6. Calculate the equilibrium concentrations of all gases if 0.800 moles of NOCL are placed in a 2.00 L container.
I ended up with NOCL^2 = 0.4 mol/L
Cl2 = 3.9 x 10^-3
NO^2 = 2x = 2 x 3.9 x 10-3 =0.0078
7.8 x 10^-3
I am hoping someone could go through the quadratic formula part and let me know if my answers are correct and if not where I took a wrong turn.
You ended up with a quadratic?
0.8mol/2L = 0.4 M
..........2NOCl2 ==> 2NO + Cl2
I.........0.40........0.....0
C..........-2x........2x....x
E.......0.40-2x.......2x....x
I ended up with a cubic equation but I found a cubic calculator on the web and it returned 0.00395 for Cl2. It appears your answers are ok.
AgCl is dissolved separatel in pure water and in O.O25 (M) NaCl solution.Find the ratio of concentrations of Ag ion in pure water and in the NaCl solution. The solubility product of AgCl is 1.75 to multiply 1O^-1O
To calculate the equilibrium concentrations of all gases, we need to use the equilibrium constant expression and solve for the unknowns. Let's go through the calculation step-by-step.
1. Given information:
- Initial moles of NOCL = 0.800 mol
- Volume of the container = 2.00 L
2. Determine the initial concentration of NOCL:
The initial concentration of NOCL can be calculated using the formula:
Initial concentration (NOCL) = moles of NOCL / volume of container
Initial concentration (NOCL) = 0.800 mol / 2.00 L = 0.400 mol/L
3. Let x be the change in concentration for both NO and Cl2. This will also be the equilibrium concentration of NO and Cl2.
4. The equilibrium concentrations of NO and Cl2 are both 2x, according to the balanced equation.
5. The equilibrium concentration of NOCL will be (0.400 - x)^2 since its coefficient is 2 in the balanced equation.
6. Write the expression for the equilibrium constant:
Kc = ([NO]^2 * [Cl2]) / [NOCL]^2
Substitute the expressions for equilibrium concentrations:
Kc = ([2x]^2 * [2x]) / [(0.400 - x)^2]^2
7. Simplify the expression:
Kc = (4x^3) / (0.1600 - 0.800x + x^2)^2
8. Substitute the given value for the equilibrium constant:
1.60 x 10^-6 = (4x^3) / (0.1600 - 0.800x + x^2)^2
9. Rearrange the equation and solve for x using the quadratic formula:
4x^3 = (1.60 x 10^-6) * (0.1600 - 0.800x + x^2)^2
(0.1600 - 0.800x + x^2)^2 = 4x^3 / (1.60 x 10^-6)
Take the square root of both sides:
0.1600 - 0.800x + x^2 = √(4x^3 / (1.60 x 10^-6))
0.1600 - 0.800x + x^2 = (2x / 0.002)^(1/2)
Expanding the right side:
0.1600 - 0.800x + x^2 = (2x / 0.002)^(1/2)
Solving for x will give you the value of the change in concentration for both NO and Cl2.
10. Once you have the value of x, substitute it back into the equilibrium expression to calculate the equilibrium concentrations of NO, Cl2, and NOCL.
Equilibrium concentration (NO) = 2x
Equilibrium concentration (Cl2) = 2x
Equilibrium concentration (NOCL) = (0.400 - x)^2
Evaluate these expressions to obtain the final equilibrium concentrations.
To solve this problem, let's start by setting up the equilibrium expression using the given equilibrium constant (Kc) value:
Kc = [NO]^2 [Cl2] / [NOCL]^2
We are given that the initial concentration of NOCL is 0.800 mol/L. Since NOCL decomposes completely, the change in concentration of NOCL will be -x, and the change in concentration of both NO and Cl2 will be +2x, assuming the stoichiometric coefficient of NOCL is 2.
Let's define our variables:
[NOCL] = 0.800 - x
[NO] = 2x
[Cl2] = 2x
Now we can substitute these values into the equilibrium expression:
Kc = (2x)^2 (2x) / (0.800 - x)^2
Simplifying this expression gives us:
1.60 x 10^-6 = 8x^3 / (0.800 - x)^2
Now, for the quadratic formula part, we need to rearrange the equation in the form of ax^2 + bx + c = 0, where:
a = 8
b = 0
c = -1.60 x 10^-6 (0.800 - x)^2
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
x = (± √(0 - 4 * 8 * (-1.60 x 10^-6 (0.800 - x)^2))) / (2 * 8)
Simplifying further,
x = ± √((-6.4 x 10^-6 (0.800 - x)^2)) / 16
x = ± √(-0.4 (0.800 - x)^2) / 16
Since the equilibrium concentration must be positive, we take the positive root:
x = √(-0.4 (0.800 - x)^2) / 16
Now, we can solve for x using trial and error or numerical methods. Let's assume x is very small compared to 0.800, so we can approximate (0.800 - x) as 0.800.
x ≈ √(-0.4 (0.800)^2) / 16
x ≈ √(-0.256) / 16
Since the square root of a negative value is undefined in real numbers, the approximation we made is not valid in this case.
To determine the equilibrium concentrations accurately, you will need to use numerical methods like solving the equation iteratively or using a graphing calculator or software.