unit thermochemistry thermochemical equations wksh 3 Doc Template ... - Free Printable

Let’s solve each problem step by step.

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Problem 1:

We are given the reaction:

C₂H₆ + 3½ O₂ → 2 CO₂ + H₂O
ΔH = -1390 kJ

This means when 1 mole of ethane (C₂H₆) burns, it releases 1390 kJ of energy (negative sign means heat is released).

We are asked: How much energy is released when 2.3 moles of ethane burn?

So we set up a proportion:

Energy released = (moles of ethane) × (energy per mole)

= 2.3 mol × (-1390 kJ/mol)

Calculate that:

2.3 × 1390 = ?

First, 2 × 1390 = 2780
Then, 0.3 × 1390 = 417
Total = 2780 + 417 = 3197

So, 2.3 × (-1390) = -3197 kJ

But wait — let’s check with exact multiplication:

2.3 × 1390 = 2.3 × (1400 - 10) = 2.3×1400 - 2.3×10 = 3220 - 23 = 3197 → same result.

So answer is -3197 kJ

But looking at the handwritten answer in the image, it says “-3200 kJ” — which is rounded to 2 significant figures? Let’s check sig figs.

The given value: 2.3 moles → has 2 sig figs
1390 kJ → has 3 or 4? Usually trailing zero without decimal is ambiguous, but in thermochemistry problems like this, often treated as exact or based on context.

However, 2.3 has 2 sig figs, so answer should have 2 sig figs.

3197 rounded to 2 sig figs is 3200, and since it’s negative: -3200 kJ

✔ So final answer for #1: -3200 kJ

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Problem 2:

Reaction:
2 CO + O₂ → 2 CO₂
ΔH = -787 kJ

This means when 2 moles of CO₂ are produced, 787 kJ of heat is released.

We want to know how many moles of CO must react to produce 147 kJ of energy.

Note: The ΔH is for producing 2 moles of CO₂, and also corresponds to reacting 2 moles of CO (from the balanced equation).

So, if 2 moles of CO → release 787 kJ

Then, x moles of CO → release 147 kJ

Set up proportion:

x / 2 = 147 / 787

Wait — actually, better to think:

Heat released per mole of CO reacted:

Since 2 moles CO → 787 kJ released
→ 1 mole CO → 787/2 = 393.5 kJ released

But we don’t need per mole — we can do direct ratio.

Moles of CO needed = (147 kJ) × (2 mol CO / 787 kJ)

Because: for every 787 kJ released, you used 2 moles of CO.

So:

moles CO = 147 × (2 / 787)

Calculate:

First, 2 / 787 ≈ 0.0025413

Then, 147 × 0.0025413 ≈ ?

Or better: 147 × 2 = 294
Then divide by 787: 294 ÷ 787

Do division:

787 goes into 294 zero times → 0.

Add decimal: 2940 ÷ 787 ≈ 3.736? Wait no:

Actually: 787 × 0.3 = 236.1
294 - 236.1 = 57.9 → bring down 0 → 579

787 × 0.07 = 55.09 → subtract → 579 - 550.9? Wait, better use calculator-style:

294 ÷ 787 = ?

Use approximation: 787 ≈ 800 → 294/800 = 0.3675

Actual: 294 ÷ 787 = 0.37357...

So approximately 0.374 moles

But let’s compute exactly:

294 ÷ 787 = ?

Do long division:

787 ) 294.000

787 > 294 → 0.

2940 ÷ 787 ≈ 3 (since 787×3=2361)

2940 - 2361 = 579

Bring down 0 → 5790

787 × 7 = 5509

5790 - 5509 = 281

Bring down 0 → 2810

787 × 3 = 2361

2810 - 2361 = 449

So far: 0.373...

So ≈ 0.3736 moles

Rounded to 3 sig figs? Given values: 147 kJ (3 sig figs), 787 kJ (3 sig figs) → so answer should have 3 sig figs.

0.374 moles

But the handwritten answer says “0.374 mol CO” — matches!

✔ Final answer for #2: 0.374 mol CO

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Problem 3:

Reaction:
CS₂ + 2 S → CS₄? Wait — typo? Probably meant CS₂ formation from elements.

Given:
C + 2 S → CS₂
ΔH = +89.3 kJ

Positive ΔH means endothermic — absorbs heat.

Question: How much heat is required to produce 6.5 moles of carbon disulfide (CS₂)?

From the equation: making 1 mole of CS₂ requires +89.3 kJ

So for 6.5 moles:

Heat required = 6.5 mol × 89.3 kJ/mol

Calculate:

6.5 × 89.3

Break it down:

6 × 89.3 = 535.8
0.5 × 89.3 = 44.65
Total = 535.8 + 44.65 = 580.45 kJ

Now, sig figs: 6.5 has 2 sig figs, 89.3 has 3 → limiting is 2 sig figs.

580.45 rounded to 2 sig figs is 580 kJ? But 580 has 2 or 3? Ambiguous.

Better to write in scientific notation: 5.8 × 10² kJ → clearly 2 sig figs.

But 580.45 rounded to 2 sig figs: look at first two digits: 5 and 8 → next digit is 0, which is less than 5? Wait:

580.45 — to 2 sig figs: start from left: 5 and 8 → then round based on third digit, which is 0 → so 580 → but that’s still 3 digits.

Actually, 580 has 2 sig figs only if written as 5.8 × 10².

In many contexts, they accept 580 as having 2 sig figs if specified.

But let’s see: 6.5 × 89.3

Compute more precisely:

6.5 × 89.3 = 6.5 × (90 - 0.7) = 6.5×90 - 6.5×0.7 = 585 - 4.55 = 580.45 → same.

Now, 6.5 has uncertainty in tenths place, 89.3 in tenths — product should have 2 sig figs.

580.45 → rounded to 2 sig figs: 580 → but that’s not standard. Actually, 5.8 × 10² = 580, which implies 2 sig figs.

Handwritten answer says “+580 kJ” — so they’re using 3 sig figs? Or accepting 580 as 2 sig figs.

Perhaps in this context, since 6.5 might be considered 2 sig figs, but 89.3 is 3, and sometimes the rule is to go with least number of sig figs in multiplication — here 6.5 has 2, so answer should have 2.

But 580.45 to 2 sig figs is 580 — which is awkward. Better to say 5.8 × 10² kJ.

However, the handwritten answer is “+580 kJ”, and likely expected.

Check calculation again: 6.5 × 89.3

Alternative: 65 × 893 / 100

65 × 893:

60×893 = 53580
5×893 = 4465
Total = 58045
Divide by 100 → 580.45 → yes.

So, 580.45 kJ → rounded to 3 sig figs is 580 kJ (if we consider 580 as having 3, but technically trailing zero without decimal is ambiguous).

Given that the problem gives 6.5 (possibly 2 sig figs) and 89.3 (3 sig figs), and the answer key shows 580, I’ll go with 580 kJ as acceptable.

Also, note: ΔH is positive, so heat absorbed → +580 kJ

✔ Final answer for #3: +580 kJ

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Problem 4:

Reaction:
2 NaHCO₃ → Na₂CO₃ + H₂O + CO₂
ΔH = +129 kJ

This means decomposing 2 moles of baking soda (NaHCO₃) requires 129 kJ of heat (endothermic).

We are given: mass of baking soda = 224 grams

Need to find heat required to decompose 224 g.

First, find molar mass of NaHCO₃:

Na = 23
H = 1
C = 12
O₃ = 48
Total = 23+1+12+48 = 84 g/mol

So, moles of NaHCO₃ = mass / molar mass = 224 g / 84 g/mol

Calculate: 224 ÷ 84

84 × 2 = 168
224 - 168 = 56
56 / 84 = 2/3 ≈ 0.6667
So total moles = 2 + 2/3 = 8/3 ≈ 2.6667 moles

Exactly: 224 / 84 = 224 ÷ 28 = 8, 84 ÷ 28 = 3 → wait, gcd of 224 and 84.

224 ÷ 28 = 8, 84 ÷ 28 = 3? 28×8=224, 28×3=84 → yes! So 224/84 = 8/3 moles

So, moles = 8/3 mol

Now, from reaction: 2 moles NaHCO₃ require 129 kJ

So, heat required for 8/3 moles = ?

Set up proportion:

(8/3 mol) × (129 kJ / 2 mol) = (8/3) × (129/2) kJ

Simplify: (8 × 129) / (3 × 2) = (8 × 129) / 6

8/6 = 4/3, so (4/3) × 129

Or: 129 / 6 = 21.5, then × 8 = 172

Compute:

129 ÷ 6 = 21.5
21.5 × 8 = 172

Yes.

(8 × 129) = 1032
1032 ÷ 6 = 172

So, heat required = 172 kJ

Sig figs: 224 g has 3 sig figs, 129 kJ has 3, molar mass 84 is exact? Usually atomic masses are considered exact in such problems, or 84 has 2 sig figs? But NaHCO₃ molar mass is typically taken as 84.0, so 3 sig figs.

224 has 3, 129 has 3 → answer should have 3 sig figs.

172 has 3 sig figs → perfect.

Handwritten answer says “172 kJ” — matches.

✔ Final answer for #4: 172 kJ

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Final Answer:
1. -3200 kJ
2. 0.374 mol CO
3. +580 kJ
4. 172 kJ