A mixture containing 0.5 mol of H2 (g) and 0.5 mol of I2 (g) was introduced into a 10L container at a temperature of 448 C. At this temperature the equilibrium constant Kp is 50.

Question

A mixture containing 0.5 mol of H2 (g) and 0.5 mol of I2 (g) was introduced into a 10L container at a temperature of 448 C. At this temperature the equilibrium constant Kp is 50.

i. What is the total pressure in the container before the reaction starts and at equilibrium.

ii. How many moles of hydrogen remain at equilibrium

iii. What is the partial pressure of each component in the equilibrium mixture

Thanks!!!

Answer 1

See your other post below.

Answer 2

When 50ml of 0.1m NH4OH is added to 50ml of 0.05ml HCL solution the pH is

Answer 3

To answer these questions, we need to use the ideal gas law and the expression for the equilibrium constant (Kp) to calculate the total pressure in the container before the reaction starts and at equilibrium, the moles of hydrogen remaining at equilibrium, and the partial pressures of each component in the equilibrium mixture.

Let's start by calculating the total pressure in the container before the reaction starts. Since the ideal gas law states that PV = nRT, we can rearrange it to solve for pressure (P):

P = (nRT) / V

Where:
P = Pressure
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin (448 °C = 721 K)
V = Volume of the container (10 L)

For the total pressure before the reaction starts, we'll consider the sum of the partial pressures of hydrogen and iodine:

P_total = P_H2 + P_I2

We can calculate each partial pressure using the equation:

P_partial = (n_partial * R * T) / V

i. Total pressure in the container before the reaction starts and at equilibrium:
Before the reaction starts, both H2 and I2 are present in equal amounts, so the partial pressure of each gas is equal. Therefore, the total pressure will be:

P_total = P_H2 + P_I2
= ((n_H2 + n_I2) * R * T) / V
= ((0.5 mol + 0.5 mol) * 0.0821 L·atm/mol·K * 721 K) / 10 L
= 29.8695 atm

At equilibrium, we'll use the equilibrium constant (Kp) to determine the partial pressures of each gas. The balanced chemical equation for the reaction is:

H2 (g) + I2 (g) ⇌ 2HI (g)

According to the expression for Kp, we have:

Kp = (P_HI^2) / (P_H2 * P_I2)

Since Kp = 50, we can rewrite this equation as:

50 = (P_HI^2) / (P_H2 * P_I2)

Since we know that the initial moles of H2 and I2 are equal, and using the stoichiometry of the reaction, we can say that at equilibrium, the moles of H2 remaining will be (0.5 - 2x), where x is the moles of HI formed.

ii. Moles of hydrogen remaining at equilibrium:
The reaction stoichiometry tells us that for every 2 moles of HI formed, 1 mole of H2 will be consumed. Therefore, we can write the expression:

(0.5 mol - 2x) = x

Simplifying the equation, we find:

0.5 mol - 3x = 0
3x = 0.5 mol
x = 0.1667 mol

So, at equilibrium, 0.1667 moles of HI are formed, and therefore, 2 * 0.1667 = 0.3334 moles of H2 remain.

iii. Partial pressure of each component in the equilibrium mixture:
We can calculate the partial pressure of each component using the ideal gas law equation:

P_partial = (n_partial * R * T) / V

For hydrogen (H2):

P_H2 = (0.3334 mol * 0.0821 L·atm/mol·K * 721 K) / 10 L
= 1.426 atm

Similarly, for iodine (I2):

P_I2 = (0.3334 mol * 0.0821 L·atm/mol·K * 721 K) / 10 L
= 1.426 atm

And for hydrogen iodide (HI), which is 2 times the moles of iodine used:

P_HI = 2 * x * R * T / V
= 2 * 0.1667 mol * 0.0821 L·atm/mol·K * 721 K / 10 L
= 2.456 atm

So, the partial pressure of hydrogen (H2) is 1.426 atm, the partial pressure of iodine (I2) is 1.426 atm, and the partial pressure of hydrogen iodide (HI) is 2.456 atm in the equilibrium mixture.