Calculus in Chemistry: Rates, Integrals, and Energy
TL;DR: Derivatives explain reaction rates; integrals compute energy and work; differential equations model how concentrations change over time. If you want pros to handle the calculus and the chemistry, we do chemistry homework—fast, private, A/B Guarantee.
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Table of Contents
1) Why Chemistry Needs Calculus
- Derivatives = change: Rates of reaction use d[A]/dt, not just average changes.
- Integrals = accumulation: Work, energy, and areas under curves use ∫.
- Differential equations = dynamics: How concentrations evolve with time.
Bridge the math & the chem
We interpret results (units, signs, assumptions), not just compute them.
2) Reaction Rates and Derivatives
For a first-order reaction A → products, the instantaneous rate is proportional to concentration:
| Statement | Math | Takeaway |
|---|---|---|
| Rate law (first order) | −d[A]/dt = k[A] | Loss of A scales with [A] |
| Given concentration vs time | [A](t) = [A]0 e−kt | Exponential decay (k > 0) |
| Differentiate | d[A]/dt = −k[A]0 e−kt = −k[A] | Derivative matches rate law |
Arrhenius tip: Plot ln k vs 1/T to get slope −Ea/R. That’s calculus + logs + thermodynamics in one graph.
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Exponents & logs show up everywhere in kinetics.
Exponents in Chemistry
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3) Integrals in Thermodynamics (PV Work)
Work for a quasi-static process is the area under the P–V curve:
W = ∫V₁→V₂ P(V) dV
Isothermal ideal gas (n moles)
For P = nRT / V at constant T,
W = ∫(nRT/V) dV = nRT ln(V₂/V₁)
Sign convention: Many chem texts take work by the system as positive for expansion (V₂ > V₁ → W > 0). Your course may use the opposite—check instructions.
Worked example
Given: n = 1.00 mol ideal gas expands isothermally at T = 298 K from V₁ = 5.00 L to V₂ = 10.00 L. R = 8.314 J·mol⁻¹·K⁻¹.
W = (1.00)(8.314)(298) ln(10.00/5.00) = 2477 · ln 2 = 1718 J (4 s.f.).
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4) Differential Equations in Kinetics
First-order integrated rate law
−d[A]/dt = k[A] ⇒ d[A]/[A] = −k dt ⇒ ∫ d[A]/[A] = − ∫ k dt
ln[A] = −kt + C ⇒ ln([A]/[A]₀) = −kt ⇒ [A] = [A]₀ e−kt
Second-order (single reactant) integrated rate law
−d[A]/dt = k[A]² ⇒ d[A]/[A]² = −k dt ⇒ ∫ [A]⁻² d[A] = − ∫ k dt
−1/[A] = −kt + C ⇒ 1/[A] = kt + 1/[A]₀
Zero-order integrated rate law
−d[A]/dt = k ⇒ [A] = [A]₀ − kt
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We fit the right model (0th/1st/2nd), plot linear forms, and extract k with units.
5) Practice Problems (with Solutions)
- Differentiate kinetics: If [A](t) = [A]₀ e−kt, compute d[A]/dt and interpret its sign.
- Isothermal work: n = 2.00 mol, T = 350 K, V₁ = 4.00 L, V₂ = 8.00 L. Compute W for an ideal gas isothermal expansion (J).
- Half-life (1st order): k = 0.115 min⁻¹. Starting at 0.800 M, what is [A] after 3 half-lives?
- Second order: For 1/[A] = 1/[A]₀ + kt with [A]₀ = 0.100 M and k = 0.250 M⁻¹·s⁻¹, find [A] at t = 12.0 s.
- Zero order: A decomposes with k = 0.0200 M·s⁻¹ from [A]₀ = 0.300 M. Time to reach 0.120 M?
Show Solutions
- d[A]/dt = −k [A]₀ e−kt = −k[A]. Negative sign shows concentration decreases for k > 0.
- W = nRT ln(V₂/V₁) = (2.00)(8.314)(350) ln(8/4) = 5810 · ln 2 = 4027 J (4 s.f.).
- For first order, t½ = ln 2 / k ≈ 0.693/0.115 = 6.03 min. After 3 half-lives, [A] = [A]₀(1/2)³ = 0.800 × 0.125 = 0.100 M.
- 1/[A] = 1/0.100 + (0.250)(12.0) = 10.0 + 3.00 = 13.0 ⇒ [A] = 0.0769 M.
- [A] = [A]₀ − kt ⇒ 0.120 = 0.300 − 0.0200 t ⇒ t = (0.180)/0.0200 = 9.00 s.
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6) Common Mistakes (and Fixes)
- Chain rule slips: d/dt (e−kt) = −k e−kt, not just e−kt.
- Mixing definite/indefinite integrals: Include limits and units for ∫ PdV.
- Sign conventions: Clarify whether expansion work is taken as + or − in your course.
- Wrong order fit: For a linear plot, try ln[A] vs t (1st order) or 1/[A] vs t (2nd order).
- Rounding too early: Carry extra digits; round at the end per instructions/sig figs.
Fix the leaks in your workflow
We set up models, pick the right plot, and check units/signs so points don’t leak away.
7) Platform Notes: ALEKS • WebAssign • MyLab
- ALEKS: Kinetics and decay modules use first/second-order models; watch scientific notation.
- WebAssign: PV-work integrals and sign conventions are enforced; include units.
- MyLab Chemistry: Rate law derivations and linearization plots (ln[A] vs t, 1/[A] vs t).
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8) How FMMC Helps
- Kinetics derivatives, PV-work integrals, and integrated rate laws—done right and on time.
- ALEKS/WebAssign/MyLab formatting and precision included.
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9) FAQs
Do I need calculus for general chemistry?
A little—mostly for understanding rates, exponentials, and areas under curves. It grows in physical chemistry and kinetics.
How does calculus show up in kinetics?
Rate laws are differential equations; integrated forms predict [A](t) and half-lives.
Why is PV work an integral?
Pressure can change with volume; the area under P(V) gives total energy transferred as work.
Can FMMC do my calculus exam if it’s chemistry-based?
Yes—see Calculus Exam Help. We also offer full class coverage.
What if my course uses ALEKS/WebAssign/MyLab?
We match the platform’s rounding, units, and formatting so your answers pass validation.
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