Answer:
3.69 atm
Step-by-step explanation:
To solve this problem, we can use an ICE (Initial, Change, Equilibrium) table and the stoichiometry of the reaction. Let's assume that the initial pressure of hydrogen iodide (HI) is 'x' atm.
The balanced chemical equation for the decomposition of hydrogen iodide is:
2 HI(g) ⇌ H2(g) + I2(g)
According to the equation, 2 moles of HI decompose to form 1 mole of H2 and 1 mole of I2. Therefore, the change in pressure for HI will be -2x, and the changes for H2 and I2 will be +x each.
Now, let's fill out the ICE table:
| 2 HI(g) ⇌ H2(g) + I2(g)
To solve this problem, we can use an ICE (Initial, Change, Equilibrium) table and the stoichiometry of the reaction. Let's assume that the initial pressure of hydrogen iodide (HI) is 'x' atm.
The balanced chemical equation for the decomposition of hydrogen iodide is:
2 HI(g) ⇌ H2(g) + I2(g)
According to the equation, 2 moles of HI decompose to form 1 mole of H2 and 1 mole of I2. Therefore, the change in pressure for HI will be -2x, and the changes for H2 and I2 will be +x each.
Now, let's fill out the ICE table:
| 2 HI(g) ⇌ H2(g) + I2(g)
Initial | x 0 0
Change | -2x +x +x
Equilibrium| x-2x x x
The equilibrium expression for the reaction is given by:
Kc = [H2] * [I2] / [HI]^2
Given that the equilibrium constant (Kc) is 1.84 x 10^-2, and the equilibrium pressures of H2 and I2 are both 0.500 atm, we can substitute these values into the equilibrium expression:
1.84 x 10^-2 = (0.500) * (0.500) / (x)^2
Simplifying this equation, we get:
1.84 x 10^-2 = 0.250 / (x^2)
To solve for x, we can rearrange the equation:
(x^2) = 0.250 / (1.84 x 10^-2)
(x^2) = 13.59
x ≈ √13.59
x ≈ 3.69
Since x represents the equilibrium pressure of HI, the equilibrium pressure of hydrogen iodide is approximately 3.69 atm.
Hope this helps!