Question

if the reaction begins with 1.00 m of h₂ and f₂ and no hf, calculate the concentration of h₂ and hf at equilibrium if the equilibrium of f₂ is 0.3 m
h₂(g) + f₂(g) <--> 2 hf (g)

	h₂	f₂	2 hf
initial
change
equilibrium

if the reaction begins with 1.40 m of h₂ and f₂ and no hf, calculate the concentration of h₂ and hf at equilibrium if the equilibrium of f₂ is 0.5 m
h₂(g) + f₂(g) <--> 2 hf (g)

	h₂	f₂	2 hf
initial
change
equilibrium

Explanation:

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First Problem:

Step1: Fill initial concentrations

Initial: $[\text{H}_2] = 1.00\ \text{M}$, $[\text{F}_2] = 1.00\ \text{M}$, $[\text{HF}] = 0\ \text{M}$

Step2: Find change in $\text{F}_2$

Change in $\text{F}_2$: $1.00 - 0.3 = 0.7\ \text{M}$

Step3: Relate changes to stoichiometry

$\Delta [\text{H}_2] = -0.7\ \text{M}$, $\Delta [\text{HF}] = +2\times0.7 = 1.4\ \text{M}$

Step4: Calculate equilibrium values

$[\text{H}_2]_{\text{eq}} = 1.00 - 0.7 = 0.3\ \text{M}$
$[\text{HF}]_{\text{eq}} = 0 + 1.4 = 1.4\ \text{M}$

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Second Problem:

Step1: Fill initial concentrations

Initial: $[\text{H}_2] = 1.40\ \text{M}$, $[\text{F}_2] = 1.40\ \text{M}$, $[\text{HF}] = 0\ \text{M}$

Step2: Find change in $\text{F}_2$

Change in $\text{F}_2$: $1.40 - 0.5 = 0.9\ \text{M}$

Step3: Relate changes to stoichiometry

$\Delta [\text{H}_2] = -0.9\ \text{M}$, $\Delta [\text{HF}] = +2\times0.9 = 1.8\ \text{M}$

Step4: Calculate equilibrium values

$[\text{H}_2]_{\text{eq}} = 1.40 - 0.9 = 0.5\ \text{M}$
$[\text{HF}]_{\text{eq}} = 0 + 1.8 = 1.8\ \text{M}$

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Answer:

First ICE Table and Results:

	$\text{H}_2$	$\text{F}_2$	$2\text{HF}$
Change	$-0.7\ \text{M}$	$-0.7\ \text{M}$	$+1.4\ \text{M}$
Equilibrium	$0.3\ \text{M}$	$0.3\ \text{M}$	$1.4\ \text{M}$

Equilibrium $[\text{H}_2] = 0.3\ \text{M}$, Equilibrium $[\text{HF}] = 1.4\ \text{M}$

Second ICE Table and Results:

	$\text{H}_2$	$\text{F}_2$	$2\text{HF}$
Change	$-0.9\ \text{M}$	$-0.9\ \text{M}$	$+1.8\ \text{M}$
Equilibrium	$0.5\ \text{M}$	$0.5\ \text{M}$	$1.8\ \text{M}$

Equilibrium $[\text{H}_2] = 0.5\ \text{M}$, Equilibrium $[\text{HF}] = 1.8\ \text{M}$