Question

The equilibrium constant for the gaseous reaction
H2 + I2 ↔ 2HI is 50.2 at 448oC, calculate the number
of grams of HI that are in equilibrium with 1.25 mole of
H2 and 63.5g of Iodine at this temperature

Answer 1

Answer on Question #44138 - Chemistry - Inorganic Chemistry

Question:

The equilibrium constant for the gaseous reaction $H_2 + I_2 \leftrightarrow 2HI$ is 50.2 at $448^{\circ}C$. Calculate the number of grams of HI that are in the equilibrium with 1.25 mole of $H_2$ and $63.5\mathrm{g}$ of iodine at this temperature.

Answer:

The constant for the equilibrium $H_2 + I_2 \leftrightarrow 2HI$ is given by the following expression:

$$K = \frac{[HI]^2}{[H_2][I_2]} = \frac{n^2(HI)}{n(H_2)n(I_2)} = 50.2$$

The amount of moles of iodine is:

$$n(I_2) = \frac{m(I_2)}{M(I_2)} = \frac{63.5\mathrm{g}}{254\mathrm{g/mol}} = 0.25\mathrm{mol}$$

Using expression for the equilibrium constant one can find the amount of moles of HI that are in the equilibrium with given amounts of $H_2$ and $I_2$:

$$50.2 = \frac{n^2(HI)}{1.25 \times 0.25};$$ $$50.2 = \frac{n^2(HI)}{0.3125};$$ $$n^2(HI) = 50.2 \times 0.3125 = 15.7$$ $$n(HI) = 3.96\mathrm{mol}$$

The mass of the HI can be calculated in the following way by multiplying the amount of moles by the molar mass:

$$m(HI) = n(HI) \times M(HI) = \frac{3.96\mathrm{mol}}{\times 128\mathrm{g/mol}} = 507\mathrm{g}.$$

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