Question 3 of 69intermediate🔧 ApplyNumerical3 marks

At a certain temperature and total pressure of 10510^5 Pa, iodine vapour contains 40% by volume of I atoms

I2(g)⇌2I(g)I_2(g) \rightleftharpoons 2I(g)

Calculate KpK_p for the equilibrium.

Correct Answer

Kp=2.67×104K_p = 2.67 \times 10^4 Pa

Exercise: EXERCISES | Q: 6.3 | (Page 41)
For More Understanding

Explanation

For gaseous mixtures, volume percentage is equal to mole fraction. Using the mole fractions, we can find the partial pressures of the gases, which are then substituted into the equilibrium constant expression Kp=(pI)2pI2K_p = \frac{(p_I)^2}{p_{I_2}} to find the value of KpK_p. The unit for KpK_p in this case is Pa, as the total pressure is given in Pa and the exponents in the numerator and denominator are not the same.

Solution Steps

  1. Step 1: Determine mole fractions For gases, volume percentage is equal to mole fraction. Given 40% by volume of I atoms: Mole fraction of I atoms, xI=0.40x_I = 0.40 Mole fraction of I2I_2, xI2=1−0.40=0.60x_{I_2} = 1 - 0.40 = 0.60

  2. Step 2: Calculate partial pressures Total pressure, Ptotal=105P_{total} = 10^5 Pa Partial pressure of I, pI=xI×Ptotal=0.40×105p_I = x_I \times P_{total} = 0.40 \times 10^5 Pa Partial pressure of I2I_2, pI2=xI2×Ptotal=0.60×105p_{I_2} = x_{I_2} \times P_{total} = 0.60 \times 10^5 Pa

  3. Step 3: Write the expression for KpK_p For the equilibrium I2(g)⇌2I(g)I_2(g) \rightleftharpoons 2I(g), the equilibrium constant in terms of partial pressure is:

    Kp=(pI)2pI2K_p = \frac{(p_I)^2}{p_{I_2}}

  4. Step 4: Substitute values and calculate KpK_p

    Kp=(0.40×105)20.60×105K_p = \frac{(0.40 \times 10^5)^2}{0.60 \times 10^5} Kp=0.16×10100.60×105K_p = \frac{0.16 \times 10^{10}}{0.60 \times 10^5} Kp=0.160.60×105K_p = \frac{0.16}{0.60} \times 10^5

    Kp=0.2667×105K_p = 0.2667 \times 10^5 Pa Kp=2.67×104K_p = 2.67 \times 10^4 Pa

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