Practice worksheet for balancing chemical equations with examples and blank lines for solutions.

Let’s solve each problem step by step. We’ll use the given equilibrium constants and rules for combining reactions.

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Problem 1:

Given:
- A ⇌ B + C → K₁ = 0.25
- C ⇌ D + E → K₂ = 0.50

We want:
A ⇌ B + D + E → K₃ = ?

Notice: If we add the two reactions together:

A ⇌ B + C
C ⇌ D + E
──────────────
A ⇌ B + D + E (C cancels out)

When you add reactions, you multiply their equilibrium constants.

So:
K₃ = K₁ × K₂ = 0.25 × 0.50 = 0.125

✔ Answer: 0.125

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Problem 2:

Given:
- 2NOBr(g) ⇌ 2NO(g) + Br₂(g) → Kc = 4.8×10⁻³

We want:
½Br₂(g) + NO(g) ⇌ NOBr(g) → Kc = ?

This is the reverse of half the original reaction.

Original:
2NOBr ⇌ 2NO + Br₂ → K = 4.8×10⁻³

Reverse it:
2NO + Br₂ ⇌ 2NOBr → K_reverse = 1 / K = 1 / (4.8×10⁻³) ≈ 208.333...

Now take half of that reaction (divide all coefficients by 2):

NO + ½Br₂ ⇌ NOBr → K_new = √(K_reverse) = √(208.333...)

Calculate:
√208.333 ≈ 14.43

But let’s do it more precisely:

K_reverse = 1 / 0.0048 = 10000 / 48 = 1250 / 6 ≈ 208.3333

√(1250/6) = √(625/3) = 25 / √3 ≈ 25 / 1.732 ≈ 14.43

Wait — actually, better to write:

K_new = [K_original]^(-1/2) because we reversed AND halved.

So:
K_new = (4.8×10⁻³)^(-1/2) = 1 / √(0.0048)

√0.0048 = √(48×10⁻⁴) = √48 × 10⁻² = 4√3 × 10⁻² ≈ 6.928 × 0.01 = 0.06928

Then 1 / 0.06928 ≈ 14.43

But let’s compute exactly:

0.0048 = 48/10000 = 12/2500 = 3/625

So √(3/625) = √3 / 25

Thus K_new = 1 / (√3 / 25) = 25 / √3 = (25√3)/3 ≈ (25×1.732)/3 ≈ 43.3 / 3 ≈ 14.43

Actually, wait — I think I messed up the exponent.

If original K = 4.8e-3 for 2NOBr ⇌ 2NO + Br₂

Then for NO + ½Br₂ ⇌ NOBr, which is reverse and half:

K_new = (K_original)^{-1/2} = (4.8×10⁻³)^{-0.5}

Compute numerically:

4.8e-3 = 0.0048

Square root of 0.0048 = ? Let's calculate:

0.0048 = 48 × 10⁻⁴ → sqrt = sqrt(48) × 10⁻² = 6.9282 × 0.01 = 0.069282

Then 1 / 0.069282 ≈ 14.43

Yes.

But let’s check with calculator-style:

(0.0048)^(-0.5) = 1 / sqrt(0.0048) = 1 / 0.069282 ≈ 14.43

So Kc ≈ 14.4

But maybe they want exact form? Or rounded?

Looking at significant figures: 4.8 has 2 sig figs → so answer should have 2.

14.43 → rounds to 14

Wait — 4.8 has two sig figs, but when taking square root and reciprocal, we keep two sig figs.

14.43 → 14 (two sig figs)? But 14 has two, yes.

Actually, 14.43 is closer to 14 than 15? No, 14.43 is 14 if rounded to two sig figs? Wait no — 14 has two sig figs, but 14.43 rounded to two sig figs is 14.

But let me double-check calculation:

sqrt(0.0048) = sqrt(4.8 * 10^-3) = sqrt(4.8)*sqrt(10^-3) = 2.19 * 0.03162 ≈ ? Better:

sqrt(0.0048) = sqrt(48/10000) = sqrt(48)/100 = 6.9282/100 = 0.069282

1 / 0.069282 = 14.433...

So approximately 14.4

With two significant figures from 4.8, we report 14

But 14.4 is three sig figs. Since 4.8 has two, and operations involve powers, typically we keep two.

However, in many textbooks, they might expect 14 or 14.4.

Let me see: 4.8 × 10⁻³ → uncertainty is about ±0.1×10⁻³, so relative error ~2%, then square root reduces error, reciprocal increases — roughly same.

I think 14 is acceptable, but let’s see what the precise value is.

Alternatively, express as fraction:

K = 4.8e-3 = 48/10000 = 12/2500 = 3/625

So K_new = (3/625)^{-1/2} = (625/3)^{1/2} = 25 / √3 = (25√3)/3

Numerically: 25 * 1.73205 / 3 = 43.30125 / 3 = 14.43375

So 14.4 if we allow three sig figs, but since input was 4.8 (two sig figs), perhaps 14

Wait — looking back at problem, other values like 0.25, 0.50 have two sig figs, so probably expect two.

But 14.43 rounded to two sig figs is 14.

However, 14 has an implied precision — sometimes written as 1.4×10¹.

But I think for this context, 14 is fine.

Wait — let me check online or standard practice.

Actually, in many similar problems, they keep it as is.

Perhaps I made a mistake.

Another way:

For reaction: 2NOBr ⇌ 2NO + Br₂, Kc = [NO]^2 [Br₂] / [NOBr]^2 = 4.8e-3

For: NO + ½Br₂ ⇌ NOBr, Kc' = [NOBr] / ([NO] [Br₂]^{1/2})

Note that Kc' = 1 / sqrt(Kc) ? Let's see:

From original Kc = [NO]^2 [Br₂] / [NOBr]^2

So [NOBr]^2 / ([NO]^2 [Br₂]) = 1/Kc

Take square root: [NOBr] / ([NO] [Br₂]^{1/2}) = 1 / sqrt(Kc)

Yes! So Kc' = 1 / sqrt(Kc) = 1 / sqrt(0.0048) = 1 / 0.069282 = 14.433

So numerically 14.4

But with two sig figs, it should be 14

However, 4.8 has two sig figs, sqrt(4.8e-3) = sqrt(0.0048) = 0.069, which has two sig figs (6.9e-2), then 1/0.069 = 14.49, which rounds to 14.

0.069 has two sig figs, 1/0.069 = 14.492... → 14 (two sig figs)

Yes.

So 14

But let's confirm with calculation:

sqrt(0.0048) = ? 0.06928, which to two sig figs is 0.069

1 / 0.069 = 14.492 → rounds to 14

Yes.

So answer is 14

But I recall that in some contexts, they might write 14.4, but per sig fig rules, it should be 14.

Looking at problem 3, they have 4.8x10^-3, and answers might expect scientific notation.

Perhaps leave as 14.

I'll go with 14

Wait — let's see the next problems for consistency.

Problem 3 has large numbers, so probably not.

I think 14 is correct.

But let me calculate exactly:

4.8e-3 = 0.0048

sqrt(0.0048) = sqrt(48/10000) = (sqrt(48))/100 = (4*sqrt(3))/100 = sqrt(3)/25

1 / (sqrt(3)/25) = 25/sqrt(3) = (25*sqrt(3))/3

sqrt(3)≈1.732, so 25*1.732=43.3, /3=14.433

So 14.4 if we use three digits, but since 4.8 has two, and it's multiplication/division, the result should have two sig figs.

14.433 rounded to two sig figs is 14.

Yes.

So 14

But I feel like it might be expected as 14.4, but I'll stick with sig fig rules.

Perhaps the problem allows three.

To be safe, let's see the answer format.

I think for now, I'll put 14

No, let's check online or think differently.

In many textbooks, for such calculations, they report as calculated.

For example, if K=4.8e-3, then 1/sqrt(K) = 1/sqrt(0.0048) = 14.43, and they might write 14 or 1.4×10^1.

But in the context, other answers are decimals, so perhaps 14.

Let's move on and come back.

Actually, I recall that for equilibrium constants, sig figs are often kept as is, but let's assume two.

I'll box 14 for now.

But let's calculate numerically:

0.0048^{-0.5} = (4.8e-3)^{-0.5} = (1000/4.8)^{0.5} = (208.333)^{0.5} = 14.433

So 14.4 if rounded to three sig figs, but 4.8 has two, so perhaps 14.

I think it's fine.

Let's look at problem 4.

Perhaps I can write it as 1.4 × 10^1, but I think for this, 14 is ok.

I'll go with 14

But let's see the answer choices or something — there are none.

Another thought: in the first problem, 0.25 and 0.50, product 0.125, which is exact.

Here, 4.8e-3, so likely expect numerical value.

Perhaps 14.4 is intended.

Let me calculate with more precision.

4.8e-3 = 0.0048

sqrt(0.0048) = sqrt(48*10^-4) = sqrt(48)*10^-2 = 6.92820323*0.01 = 0.0692820323

1/0.0692820323 = 14.43375673

So approximately 14.4

And since 4.8 has two sig figs, but it's 4.8, which is between 4.75 and 4.85, so K could be from 4.75e-3 to 4.85e-3

sqrt(4.75e-3) = sqrt(0.00475) = 0.06892, 1/0.06892 = 14.51

sqrt(4.85e-3) = sqrt(0.00485) = 0.06964, 1/0.06964 = 14.36

So range from 14.36 to 14.51, so 14.4 is reasonable, and 14 is within, but 14.4 is better.

Typically in such problems, they expect 14.4 or 14.

I think I'll go with 14.4 as it's more accurate, and perhaps the sig fig rule is relaxed for logs and roots.

In many sources, for K calculations, they keep the digits.

For example, if K=4.8e-3, then 1/sqrt(K) = 14.4

So I'll put 14.4

Yes.

So for problem 2: 14.4

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Problem 3:

Given:
- N₂O₄(g) ⇌ 2NO₂(g) → Kp = 0.148 at 298 K

We want:
14NO₂(g) ⇌ 7N₂O₄(g) → Kp = ?

First, note that this is the reverse of 7 times the original reaction.

Original: N₂O₄ ⇌ 2NO₂, Kp = 0.148

Reverse: 2NO₂ ⇌ N₂O₄, Kp_reverse = 1 / 0.148

Now, multiply by 7: 14NO₂ ⇌ 7N₂O₄, so Kp_new = (Kp_reverse)^7 = (1 / 0.148)^7

First, calculate 1 / 0.148

0.148 = 148/1000 = 37/250

1 / 0.148 = 1000 / 148 = 250 / 37 ≈ 6.756756...

So approximately 6.7568

Now raise to 7th power: (6.7568)^7

That's big, but let's compute step by step.

First, 6.7568^2 = ?

6.7568 * 6.7568

6.75^2 = 45.5625, but let's calculate:

6.7568 * 6.7568

Approximately: 6.76^2 = (7-0.24)^2 = 49 - 2*7*0.24 + (0.24)^2 = 49 - 3.36 + 0.0576 = 45.6976

More accurately:

6.7568 * 6.7568

Let me compute: 6.7568 * 6 = 40.5408

6.7568 * 0.7 = 4.72976

6.7568 * 0.056 = 6.7568 * 0.05 = 0.33784, 6.7568 * 0.006 = 0.0405408, total 0.3783808

6.7568 * 0.0008 = very small, ignore for now.

Better: 6.7568^2 = (6 + 0.7568)^2 = 36 + 2*6*0.7568 + (0.7568)^2 = 36 + 9.0816 + 0.57274624 = 45.65434624

So approximately 45.654

Now ^4 = (45.654)^2

45.654^2 = (45 + 0.654)^2 = 2025 + 2*45*0.654 + (0.654)^2 = 2025 + 58.86 + 0.427716 = 2084.287716

So approximately 2084.29

Now ^6 = ^4 * ^2 = 2084.29 * 45.654 ≈ ?

First, 2000*45.654 = 91308

84.29*45.654 ≈ 84*45.654 = 84*45 = 3780, 84*0.654=54.936, total 3834.936

0.29*45.654≈13.24

So total approx 91308 + 3834.936 + 13.24 = 95156.176

This is messy.

Use calculator in mind or approximate.

Note that 0.148 is given, so 1/0.148 = 1000/148 = 250/37 ≈ 6.756756756...

Let me use 6.75676

Now (6.75676)^7

Or better, since Kp_new = (1/0.148)^7 = 1 / (0.148)^7

Compute (0.148)^7

First, 0.148^2 = 0.021904

0.148^4 = (0.021904)^2 = 0.000479785216

0.148^6 = 0.148^4 * 0.148^2 = 0.000479785216 * 0.021904 ≈ ?

0.000479785216 * 0.02 = 9.59570432e-6

0.000479785216 * 0.001904 = approximately 0.000479785216 * 0.0019 = 9.115919104e-7, and 0.000479785216 * 0.000004 = very small, so total approx 9.5957e-6 + 9.116e-7 = 1.05073e-5

Better:

0.000479785216 * 0.021904 = let's compute:

479785.216e-9 * 21904e-6 = too messy.

0.000479785216 * 0.021904 = 4.79785216e-4 * 2.1904e-2 = 1.0509e-5 approximately

Calculate:

4.79785216 * 2.1904 * 10^{-6} = first 4.79785216 * 2.1904

4.8 * 2.19 = 10.512, but more accurately:

4.79785216 * 2 = 9.59570432

4.79785216 * 0.19 = 0.9115919104

4.79785216 * 0.0004 = 0.001919140864

Sum: 9.59570432 + 0.9115919104 = 10.5072962304 + 0.001919140864 = 10.509215371264

So approximately 10.5092 * 10^{-6} = 1.05092e-5

So 0.148^6 ≈ 1.05092e-5

Then 0.148^7 = 0.148^6 * 0.148 = 1.05092e-5 * 0.148 = 1.5553616e-6

So Kp_new = 1 / (0.148)^7 = 1 / 1.5553616e-6 = 6.429e5 approximately

Calculate: 1 / 1.5553616e-6 = 10^6 / 1.5553616 ≈ 642,900

More accurately: 1,000,000 / 1.5553616 ≈ ?

1.5553616 * 642,900 = ? Or divide.

1,000,000 ÷ 1.5553616

First, 1.5553616 * 642,000 = 1.5553616 * 600,000 = 933,216.96, 1.5553616 * 42,000 = 65,325.1872, total 998,542.1472

Subtract from 1,000,000: 1,457.8528

Now 1.5553616 * 937 = 1.5553616 * 900 = 1,399.82544, 1.5553616 * 37 = 57.5483792, total 1,457.3738192

Close to 1,457.8528, difference 0.4789808

Then 1.5553616 * 0.308 ≈ 0.479, close.

So total 642,000 + 937 + 0.308 = 642,937.308

So approximately 642,937

But this is for 0.148^7, and we have 1/that.

Earlier I had 0.148^7 = 1.5553616e-6, so 1/1.5553616e-6 = 642,937

But let's use the inverse directly.

Kp_new = (1/0.148)^7

1/0.148 = 1000/148 = 250/37 ≈ 6.756756756756...

Let me use 6.756756756756

Now (6.756756756756)^7

Or better, since 250/37, so (250/37)^7 = 250^7 / 37^7

But that's huge.

Numerically, 6.756756756756^2 = 45.65432098765432 (since (250/37)^2 = 62500/1369)

62500 ÷ 1369 ≈ 45.65432098765432

Then ^4 = (45.65432098765432)^2 = 2084.287654320987 (approximately)

Calculate: 45.65432098765432 * 45.65432098765432

Or use (250/37)^4 = 250^4 / 37^4 = 3906250000 / 1874161

250^2=62500, 250^4=62500^2=3,906,250,000

37^2=1369, 37^4=1369^2=1,874,161

So 3,906,250,000 / 1,874,161 ≈ ?

Divide: 3,906,250,000 ÷ 1,874,161

First, 1,874,161 * 2084 = ? This is tedious.

From earlier calculation, we had approximately 2084.287

Then ^6 = ^4 * ^2 = 2084.287 * 45.65432 ≈ 95,156. something

Earlier I had around 95,156

Then ^7 = ^6 * 6.756756 ≈ 95,156 * 6.756756

95,000 * 6.756756 = 641,891.82

156 * 6.756756 ≈ 1,054.05336

Total approximately 642,945.87336

So Kp_new ≈ 642,946

But this is for (1/0.148)^7, and 0.148 has three sig figs, so answer should have three.

642,946 is 6.43e5

Let's use the initial approach.

0.148^7 = (0.148^2)^3 * 0.148 = (0.021904)^3 * 0.148

0.021904^2 = 0.000479785216

0.000479785216 * 0.021904 = as before ~1.05092e-5

Then * 0.148 = 1.5553616e-6

1 / 1.5553616e-6 = 642,937.5

So 6.43 × 10^5

With three sig figs, since 0.148 has three, so 6.43 × 10^5

Yes.

So Kp = 6.43e5

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Problem 4:

Given:
- N₂(g) + O₂(g) ⇌ 2NO(g) → Kp = 4.8×10⁻³¹

We want:
½N₂(g) + ½O₂(g) ⇌ NO(g) → Kp = ?

This is half of the original reaction.

When you halve the coefficients, you take the square root of K.

So Kp_new = sqrt(Kp_original) = sqrt(4.8×10⁻³¹)

sqrt(4.8×10⁻³¹) = sqrt(4.8) × sqrt(10⁻³¹) = sqrt(4.8) × 10^{-15.5}

10^{-15.5} = 10^{-15} × 10^{-0.5} = 10^{-15} / sqrt(10) ≈ 10^{-15} / 3.16227766 ≈ 3.16227766e-16

Better: sqrt(10^{-31}) = 10^{-31/2} = 10^{-15.5} = 3.16227766 × 10^{-16} ? No

10^{-15.5} = 10^{-15 - 0.5} = 10^{-15} × 10^{-0.5} = 10^{-15} × 0.316227766 ≈ 3.16227766 × 10^{-16}

10^{-0.5} = 1/sqrt(10) ≈ 0.316227766, so 3.16227766 × 10^{-1} × 10^{-15} = 3.16227766 × 10^{-16}

Yes.

Now sqrt(4.8) ≈ 2.19089

So Kp_new = 2.19089 × 3.16227766 × 10^{-16} ≈ ?

First, 2.19089 * 3.16227766 ≈ 2.19 * 3.162 = 6.92478, more accurately:

2.19089 * 3 = 6.57267

2.19089 * 0.16227766 ≈ 2.19089 * 0.16 = 0.3505424, 2.19089 * 0.00227766 ≈ 0.00499, total 0.3555324

Sum 6.57267 + 0.3555324 = 6.9282024

So approximately 6.928 × 10^{-16}

But let's compute sqrt(4.8e-31) = sqrt(4.8)*10^{-15.5} = sqrt(4.8)/10^{15.5}

10^{15.5} = 10^{15} * 10^{0.5} = 10^{15} * sqrt(10) ≈ 3.16227766e15

sqrt(4.8) = sqrt(48/10) = (sqrt(48))/sqrt(10) = (4*sqrt(3))/sqrt(10) = 4*1.73205/3.16227766 ≈ 6.9282/3.16227766 ≈ 2.19089, same as before.

So 2.19089 / 3.16227766e15 = ? No:

Kp_new = sqrt(4.8e-31) = sqrt(4.8) * sqrt(1e-31) = 2.19089 * 1e-15.5

1e-15.5 = 10^{-15.5} = 3.16227766e-16? No:

10^{-15.5} = 10^{-15 - 0.5} = 10^{-15} * 10^{-0.5} = 10^{-15} * 0.316227766 = 3.16227766 * 10^{-16}

Yes.

So 2.19089 * 3.16227766 * 10^{-16} = as above ~6.928 * 10^{-16}

But 2.19089 * 3.16227766 = let's calculate:

2.19089 * 3.16227766

First, 2.19089 * 3 = 6.57267

2.19089 * 0.16227766 = calculate 2.19089 * 0.16 = 0.3505424, 2.19089 * 0.00227766 = approximately 0.004990, total 0.3555324

Sum 6.57267 + 0.3555324 = 6.9282024

So 6.9282024 × 10^{-16}

Now, 4.8 has two sig figs, so sqrt should have two sig figs.

6.928e-16 rounded to two sig figs is 6.9e-16

Because 6.9 has two sig figs.

6.928 is closer to 6.9 than to 7.0? 6.9 to 7.0, 6.928 is greater than 6.95? No, 6.928 < 6.95, so rounds to 6.9.

Sig figs for 6.928e-16: the number is 0.0000000000000006928, so first two digits are 6 and 9, so 6.9e-16.

Yes.

So 6.9 × 10^{-16}

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Problem 5:

Given:
- 2NO₂(g) ⇌ N₂O₄(g) → Kp = 6.7 at 298 K

We want:
2N₂O₄(g) ⇌ 4NO₂(g) → Kp = ?

First, note that this is the reverse of twice the original reaction.

Original: 2NO₂ ⇌ N₂O₄, Kp = 6.7

Reverse: N₂O₄ ⇌ 2NO₂, Kp_reverse = 1 / 6.7

Now, multiply by 2: 2N₂O₄ ⇌ 4NO₂, so Kp_new = (Kp_reverse)^2 = (1 / 6.7)^2

Compute 1 / 6.7 ≈ 0.149253731

Then (0.149253731)^2 ≈ 0.022276

More accurately: 1/6.7 = 10/67 ≈ 0.14925373134328358

Square: (10/67)^2 = 100 / 4489

4489 ÷ 100 = 44.89, so 100/4489 ≈ 0.022276

Calculate: 4489 * 0.022276 = ? Or 100 ÷ 4489

4489 * 0.022 = 98.758, subtract from 100: 1.242

4489 * 0.000276 = approximately 1.239, close.

So 0.022276

With two sig figs from 6.7, so 0.022

6.7 has two sig figs, 1/6.7 = 0.149, which to three digits, but squared.

1/6.7 ≈ 0.149, (0.149)^2 = 0.022201, so 0.022

Or 2.2 × 10^{-2}

Yes.

So 0.022 or 2.2 × 10^{-2}

Probably 0.022

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Now summarize:

1. 0.125

2. 14.4

3. 6.43 × 10^5

4. 6.9 × 10^{-16}

5. 0.022

For problem 2, I decided on 14.4

For problem 5, 0.022

Now write final answers.

Also, problem 6 and 7 are not to be solved, as per instruction "Solve the following problems" and only 1-5 are listed with blanks.

In the image, problems 1-5 have blanks, 6 and 7 are separate.

So only 1-5.

Final Answer:

1. 0.125
2. 14.4
3. 6.43 × 10^5
4. 6.9 × 10^{-16}
5. 0.022

For problem 3, I have 6.43e5, but let's confirm the calculation once more.

Kp_original = 0.148 for N2O4 ⇌ 2NO2

For 14NO2 ⇌ 7N2O4, which is reverse and 7 times.

So Kp_new = (1 / Kp_original)^7 = (1/0.148)^7

1/0.148 = 6.756756756...

6.756756756^7

Or (1000/148)^7 = (250/37)^7

250^7 = 6103515625000000? Too big.

Numerically: 0.148^7 = 0.148^6 * 0.148

0.148^2 = 0.021904

0.148^4 = 0.021904^2 = 0.000479785216

0.148^6 = 0.000479785216 * 0.021904 = let's calculate exactly:

0.000479785216 * 0.021904 = 479785.216e-9 * 21904e-6 = better:

4.79785216e-4 * 2.1904e-2 = 1.050921537024e-5

Then 0.148^7 = 1.050921537024e-5 * 0.148 = 1.55536387479552e-6

Then 1 / 1.55536387479552e-6 = 642,935.5

So 6.429355e5, which to three sig figs is 6.43e5

Yes.

Similarly for others.

So final answers.