Two moles of hydrogen and one mole of iodine are taken in a 1dm3 steel at 490°c. Calculate the concentration of the gases at equilibrium.take equilibrium constant K as √2107

Answer 1

To find the concentration of the gases at equilibrium, we need to use the ideal gas law and the concept of mole ratios.

Firstly, let's find the initial number of moles for hydrogen and iodine.

For hydrogen, since we have two moles present initially, the initial number of moles of hydrogen is 2 moles.
For iodine, since we have one mole present initially, the initial number of moles of iodine is 1 mole.

Next, we need to consider the reaction between hydrogen and iodine:

H2(g) + I2(g) ⇌ 2HI(g)

From the balanced equation, we can see that stoichiometrically, one mole of iodine reacts with one mole of hydrogen to form two moles of hydrogen iodide. Thus, at equilibrium, if "x" moles of hydrogen iodide are formed, "x" moles of hydrogen and iodine will react to form "2x" moles of hydrogen iodide.

Since we have two moles of hydrogen initially, at equilibrium, the number of moles of hydrogen will be 2 - x.
Similarly, since we have one mole of iodine initially, at equilibrium, the number of moles of iodine will be 1 - x.
And, since we have no moles of hydrogen iodide initially, at equilibrium, the number of moles of hydrogen iodide will be x.

Now, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin.

At equilibrium, we assume the volume remains constant at 1 dm3 and the gases behave ideally.

For hydrogen, we have:
P(H2) = (2 - x) RT/V

For iodine, we have:
P(I2) = (1 - x) RT/V

For hydrogen iodide, we have:
P(HI) = x RT/V

Now, to relate the partial pressures of the gases to the equilibrium constant (K), we can write:

K = √(P(HI) / (P(H2) * P(I2)))

Given that K = √2107, we can substitute the expressions for the partial pressures into the equation for K:

√2107 = √((x RT/V) / ((2 - x) RT/V * (1 - x) RT/V))

Simplifying, we get:

√2107 = √((x) / ((2 - x) * (1 - x)))

Squaring both sides, we have:

2107 = (x) / ((2 - x) * (1 - x))

Expanding the denominator, we have:

2107 = (x) / (2 - 3x + x^2)

Multiplying both sides by (2 - 3x + x^2), we get:

2107(2 - 3x + x^2) = x

Expanding and rearranging, we have:

4214 - 6321x + 2107x^2 = x

Rearranging, we have the quadratic equation:

2107x^2 - 6322x + 4214 = 0

Solving this quadratic equation for x using the quadratic formula, we find two possible values for x:

x = 0.282 or x = 1.585

However, since we are interested in the concentration of gases, we need to use the value of x that is between 0 and 1, representing the fractional amount of reactant converted to product.

Therefore, the concentration of hydrogen iodide at equilibrium is x = 0.282 moles/dm3.

Using this value, we can find the concentration of hydrogen and iodine at equilibrium:

Concentration of hydrogen = 2 - x = 2 - 0.282 = 1.718 moles/dm3
Concentration of iodine = 1 - x = 1 - 0.282 = 0.718 moles/dm3

Thus, the concentration of the gases at equilibrium are as follows:
Hydrogen: 1.718 moles/dm3
Iodine: 0.718 moles/dm3
Hydrogen iodide: 0.282 moles/dm3