Question

At \(1\,\mathrm{K}\), \(15\) moles of \(\mathrm{H_2}\) reacts with \(5.2\) moles of \(\mathrm{I_2}\) and forms \(10\) moles of \(\mathrm{HI}\). The equilibrium constant for the reaction \[ 2\mathrm{HI}(g)\rightleftharpoons \mathrm{H_2}(g)+\mathrm{I_2}(g) \] is

• A. \(50\)
• B. \(100\)
• C. \(2\times10^{-2}\)
• D. \(1\times10^{-2}\) \bigskip

Hint:

Use an ICE (Initial--Change--Equilibrium) table for equilibrium problems. If all species are in the same volume, moles may be used directly in the equilibrium expression because the volume terms cancel.

Answer 1

The Correct Option is C

Concept: For reaction 2HI ⇌ H₂ + I₂, set up ICE table, solve for equilibrium moles, compute \(K_c\).

Step 1: Initial: H₂=15, I₂=5.2, HI=0. 10 mol HI formed → change in H₂/I₂ = 5. Equilibrium: H₂=10, I₂=0.2, HI=10. \(K_c = (10\times0.2)/10^2 = 2\times10^{-2}\).

Step 2: Write the final answer. \(\boxed{K_c=2\times10^{-2}}\)