03 Equilibrium Constant Worksheet 1 Answers - SCH4U1 Name: ET06 ... - Free Printable

Let’s solve problem #2 step by step.

We are given the reaction:

H₂(g) + Br₂(g) ⇌ 2 HBr(g) + energy

And we’re told:

- The equilibrium constant, K_eq = 1.02 at 25°C
- At equilibrium, [HBr] = 0.50 mol/L
- We assume that [H₂] = [Br₂] at equilibrium (they are present in equal amounts)
- We need to find [H₂] at equilibrium

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Step 1: Write the equilibrium expression

For the reaction:
H₂ + Br₂ ⇌ 2 HBr

The equilibrium constant expression is:

> K_eq = [HBr]² / ([H₂] × [Br₂])

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Step 2: Plug in what we know

We know:

- K_eq = 1.02
- [HBr] = 0.50 → so [HBr]² = (0.50)² = 0.25
- Let [H₂] = x → then [Br₂] = x (since they are equal)

So plug into the formula:

> 1.02 = 0.25 / (x × x)
> → 1.02 = 0.25 / x²

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Step 3: Solve for x²

Multiply both sides by x²:

> 1.02 × x² = 0.25

Now divide both sides by 1.02:

> x² = 0.25 / 1.02

Calculate that:

> 0.25 ÷ 1.02 ≈ 0.2451

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Step 4: Solve for x

Take square root of both sides:

> x = √0.2451 ≈ 0.495

Round to two significant figures (since 0.50 and 1.02 both have 2 or 3 — but 0.50 has 2, so we’ll go with 2):

> x ≈ 0.50

Wait — let’s check if this makes sense.

If x = 0.50, then:

K_eq = (0.50)² / (0.50 × 0.50) = 0.25 / 0.25 = 1.00

But we were told K_eq = 1.02, not 1.00.

So our answer should be *slightly less* than 0.50 to make the denominator smaller → which makes K_eq larger.

Let’s calculate more precisely:

x² = 0.25 / 1.02 = 0.245098...

x = √0.245098 ≈ 0.49507

Rounded to two decimal places? But 0.50 was given with two significant figures. However, since 1.02 has three, maybe we keep three?

Actually, looking back — the worksheet says the answer is [H₂] = 0.50 mol/L

That suggests they rounded it.

Let’s test x = 0.495:

[H₂] = 0.495, [Br₂] = 0.495

Denominator = 0.495 × 0.495 = 0.245025

Numerator = 0.25

K_eq = 0.25 / 0.245025 ≈ 1.0203 → very close to 1.02!

So technically, [H₂] ≈ 0.495 mol/L

But the worksheet says 0.50 mol/L — probably because they expect rounding to match the precision of the given [HBr] = 0.50 (which has two significant figures).

In many high school contexts, they accept 0.50 as the answer here.

Also, note: if you use exact fractions:

x² = 0.25 / 1.02 = 25/102

x = √(25/102) = 5 / √102 ≈ 5 / 10.0995 ≈ 0.495

Still ~0.495

But since the worksheet already shows the answer as 0.50 mol/L, and it's an answer key, we can conclude they want us to report 0.50 mol/L — likely due to rounding conventions or simplification.

Alternatively, perhaps there was a typo in the problem? If K_eq were exactly 1.00, then [H₂] would be exactly 0.50.

But since K_eq is 1.02, strictly speaking, [H₂] is about 0.495.

However, for consistency with the provided answer key, and since 0.495 rounds to 0.50 when using two significant figures (because 0.50 implies ±0.005), we can say:

> [H₂] = 0.50 mol/L

is acceptable in this context.

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Final Answer:
0.50 mol/L