a) Hydrogen reacts with iodine to form hydrogen iodide. \[ H_{2}(g)+I_{2}(g) \rightleftharpoons 2 H I(g) \] When \( 43.0 \mathrm{~g} \) of iodine and \( 0.8 \mathrm{~g} \) of hydrogen were heated until equilibrium was established at \( 470^{\circ} \mathrm{C} \), the equilibrium mixture was found to contain \( 1.7 \mathrm{~g} \) of iodine. Determine the equilibrium constant for this reaction. (4 marks)

          a) Hydrogen reacts with iodine to form hydrogen iodide.
\[
H_{2}(g)+I_{2}(g) \rightleftharpoons 2 H I(g)
\]

When \( 43.0 \mathrm{~g} \) of iodine and \( 0.8 \mathrm{~g} \) of hydrogen were heated until equilibrium was established at \( 470^{\circ} \mathrm{C} \), the equilibrium mixture was found to contain \( 1.7 \mathrm{~g} \) of iodine. Determine the equilibrium constant for this reaction.
(4 marks)

a) Hydrogen reacts with iodine to form hydrogen iodide.

H2(g)+I2(g) ⇌ 2 H I(g)

When 43.0 g of iodine and 0.8 g of hydrogen were heated until equilibrium was established at 470^∘C, the equilibrium mixture was found to contain 1.7 g of iodine. Determine the equilibrium constant for this reaction.
(4 marks)

Added by Luis S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Dominador Tan

Step 1

- The molar mass of \(H_2\) is approximately \(2.02 \, \text{g/mol}\). - The molar mass of \(I_2\) is approximately \(253.8 \, \text{g/mol}\). So, for hydrogen: \[0.8 \, \text{g} \, H_2 \times \frac{1 \, \text{mol} \, H_2}{2.02 \, \text{g} \, H_2} \approx 0.396 Show more…

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a) Hydrogen reacts with iodine to form hydrogen iodide. \[ H_{2}(g)+I_{2}(g) \rightleftharpoons 2 H I(g) \] When \( 43.0 \mathrm{~g} \) of iodine and \( 0.8 \mathrm{~g} \) of hydrogen were heated until equilibrium was established at \( 470^{\circ} \mathrm{C} \), the equilibrium mixture was found to contain \( 1.7 \mathrm{~g} \) of iodine. Determine the equilibrium constant for this reaction. (4 marks)