An equilibrium mixture of H_(2(g))+F_(2(g))<=>2HF_((g)) at 298 K is 0.0500MH_(2),0.0100MF_(2)&0.400MHF, all in a 5.00 L flask. a. Determine the

Part A: Calculate K_c at 298 K

Given:

• [H₂] = 0.0500 M
• [F₂] = 0.0100 M
• [HF] = 0.400 M

The equilibrium expression for the reaction:

K_c = [HF]² / ([H₂][F₂])

Substitute the known values:

K_c = (0.400)² / (0.0500 × 0.0100) = 0.1600 / 0.0005 = 320

Final Answer: K_c = 320

Part B: New Equilibrium After Adding F₂

Step 1: Determine new initial conditions

Added 0.200 mol F₂ in a 5.00 L flask:

[F₂] = 0.0100 + (0.200 mol / 5.00 L) = 0.0100 + 0.0400 = 0.0500 M

New initial concentrations:

• [H₂] = 0.0500 M
• [F₂] = 0.0500 M
• [HF] = 0.400 M

Step 2: Set up ICE table

Step 3: Plug into equilibrium expression

K_c = 320 = (0.400 + 2x)² / [(0.0500 – x)(0.0500 – x)]

This leads to:

(0.400 + 2x)² = 320 × (0.0500 – x)²

Take square root of both sides:

0.400 + 2x = √320 × (0.0500 – x)

√320 ≈ 17.89 →

0.400 + 2x = 17.89(0.0500 – x)

0.400 + 2x = 0.8945 – 17.89x

19.89x = 0.4945 → x ≈ 0.0249 M

Step 4: Final equilibrium concentrations

• [H₂] = 0.0500 – 0.0249 = 0.0251 M
• [F₂] = 0.0500 – 0.0249 = 0.0251 M
• [HF] = 0.400 + 2×0.0249 = 0.4498 M

Part C: Explanation of Concentration Changes

According to Le Châtelier’s Principle:

• Adding F₂ disturbs equilibrium by increasing a reactant.
• The system shifts right (forward reaction) to consume added F₂.
• This results in the decrease of H₂ and increase of HF.

This shift restores equilibrium by offsetting the increase in F₂, forming more HF.