Hydrogen iodide can decompose into hydrogen and iodine gases.
2HI(g) mc026-1.jpg H2(g) + I2(g)
K for the reaction is 0.016. If 0.148 atm of HI(g) is sealed in a flask, what is the pressure of each gas when equilibrium is established?
To solve this problem, we need to use the equation for the equilibrium constant (K) and the given initial pressure of HI.
The equation for the equilibrium constant (K) is given by:
K = [H2][I2] / [HI]^2
Given that K = 0.016, we'll use this information to find the equilibrium pressure of H2 and I2.
Let's assume that at equilibrium, the pressure of H2 is "x" atm, and the pressure of I2 is also "x" atm. The pressure of HI would be the initial pressure minus "x" atm, since it decomposes into H2 and I2.
So, the expression for the equilibrium constant K can be rewritten as:
0.016 = (x * x) / (0.148 - x)^2
To solve for "x," we can rearrange the equation and solve for x using the quadratic formula:
x^2 = 0.016 * (0.148 - x)^2
Expanding and rearranging the equation, we get:
0.016 * (0.148 - x)^2 - x^2 = 0
Simplifying further:
0.002368 - 0.04736x + 0.016x^2 - x^2 = 0
0.015x^2 - 0.04736x + 0.002368 = 0
Now, we can solve this quadratic equation. Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 0.015, b = -0.04736, and c = 0.002368.
Calculating the discriminant:
√(b^2 - 4ac) = √((-0.04736)^2 - 4 * 0.015 * 0.002368) = 0.042129
x = (-(-0.04736) ± 0.042129) / 2 * 0.015
Simplifying further:
x = (0.04736 ± 0.042129) / 0.03
There will be two solutions for this, one with the positive sign and one with the negative sign:
1. x = (0.04736 + 0.042129) / 0.03 = 1.6343
2. x = (0.04736 - 0.042129) / 0.03 = 0.14637
Since the pressure cannot be negative, we disregard the second solution.
Therefore, the pressure of H2 when equilibrium is established is approximately 1.6343 atm. Since the pressure of I2 is also assumed to be the same, the pressure of I2 when equilibrium is established is approximately 1.6343 atm. And the pressure of HI can be calculated as the initial pressure minus the pressure of H2:
Pressure of HI = Initial pressure of HI - Pressure of H2 = 0.148 atm - 1.6343 atm = -1.4863 atm
Since the pressure cannot be negative, we disregard this value as well.
So, the pressure of each gas when equilibrium is established is approximately:
Pressure of H2 = 1.6343 atm
Pressure of I2 = 1.6343 atm
Pressure of HI = 0 atm (disregarded the negative value)
Note: The negative value for the pressure of HI indicates that the reaction has gone to completion, and all the HI has decomposed into H2 and I2.
To determine the pressure of each gas when equilibrium is established, we need to use the equilibrium constant (K) and the initial pressure of HI(g).
The given balanced equation for the reaction is:
2HI(g) ↔ H2(g) + I2(g)
The equilibrium constant expression for this reaction is:
K = [H2] * [I2] / [HI]^2
Given that K = 0.016, we can set up the following equation:
0.016 = [H2] * [I2] / [HI]^2
Since we are given the initial pressure of HI(g) as 0.148 atm, we can substitute the value into the equation:
0.016 = [H2] * [I2] / (0.148^2)
To solve for the pressure of each gas, we need one more piece of information: the initial pressure of H2(g) and I2(g). Without this information, we cannot determine the pressure of each gas at equilibrium.
Once we have the initial pressures for H2(g) and I2(g), we can substitute the values into the equation and solve for [H2] and [I2]. The sum of the pressures of H2(g) and I2(g) will then give us the total pressure when equilibrium is established.
........2HI ==> H2 + I2
I......0.148.....0....0
C........-2x.....x....x
E.....0.148-2x...x.....x
Substitute the E line into K expression and solve for x.