FE Chemical Practice Question Solution included after investigation
CASE E-069 · FE

The Rate Constant That Increased Sixfold at 330 K

A 30 K temperature rise did not cause a 10% kinetic change. The Arrhenius relation increased the rate constant by a factor of about 6.19.

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Case E-069 engineering failure visual
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2

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3

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4

Fix

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5

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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A reaction has k1=0.10 min^-1 at 300 K and activation energy Ea=50 kJ/mol. Estimate k2 at 330 K using R=8.314 J/(mol·K). A note assumes the 10% temperature increase gives k2=0.11 min^-1. Determine the correct rate constant.

Chemical Reaction Engineering Chemical Design Verification Root Cause Analysis Safe Correction
E-069
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CASE CLOSED · ENGINEERING VERDICT

What actually failed — and what should change.

What failed

A linear percentage-temperature rule was applied to reaction kinetics.

What was missed

Rate constants vary exponentially with reciprocal absolute temperature.

Corrected model

Use the Arrhenius ratio with kelvin and consistent Ea/R units.

QUICK ANSWER

Use ln(k2/k1)=-(Ea/R)(1/T2-1/T1). With k1=0.10 min^-1 at 300 K, Ea=50,000 J/mol, and T2=330 K, k2/k1≈6.1867, so k2≈0.6187 min^-1.

ENGINEERING TAKEAWAY

Use ln(k2/k1)=-(Ea/R)(1/T2-1/T1) with Ea and R in consistent units and T in kelvin.

CASE QUESTIONS

Questions this case should settle.

Why must temperature be in kelvin in the Arrhenius equation?

The equation uses reciprocal absolute temperature, so an absolute temperature scale is required.

What is the first-order half-life when k=0.6187 min^-1?

About 1.12 minutes.

NEXT FE PRACTICE QUESTION

The 70% Conversion That Produced Only 50% Yield

Chemical Reaction Engineering · 5–10 min

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