Question
Most chemical reactions require activation energies ranging between 40 and $400 \mathrm{kJ} \mathrm{mol}^{-1} .$ What are the equivalents of 40 and $400 \mathrm{kJ} \mathrm{mol}^{-1}$ in terms of $(a) \mathrm{nm},(b)$ wave numbers, and $(c)$ electron volts?
Step 1
We use the equation $E=h \cdot c / \lambda$, where $E$ is the energy, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Solving for $\lambda$ gives us $\lambda=h \cdot c / E$. Show more…
Show all steps
Your feedback will help us improve your experience
Ernest Nachaki and 58 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The activation energies of two reactions are $18 \mathrm{~kJ}$ $\mathrm{mol}^{-4}$ and $4.0 \mathrm{~kJ} \mathrm{~mol}^{-4}$ respectively. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants at $27^{\circ} \mathrm{C}$ is (a) $3.656 \times 10^{-3}$ (b) $3.624 \times 10^{-6}$ (c) $36.52 \times 10^{-4}$ (d) $4.656 \times 10^{-4}$
The activation energies of two reactions are $18 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $4.0 \mathrm{~kJ} \mathrm{~mol}^{-4}$ respectively. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants at $27^{\circ} \mathrm{C}$ is: (a) $3.656 \times 10^{-3}$ (b) $3.624 \times 10^{-6}$ (c) $36.52 \times 10^{-8}$ (d) $4.656 \times 10^{-4}$
The activation energy $E_{\mathrm{a}}$ is $139.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ for the gasphase reaction $$ \mathrm{HI}+\mathrm{CH}_{3} \mathrm{I} \longrightarrow \mathrm{CH}_{4}+\mathrm{I}_{2} $$ Calculate the fraction of the molecules whose collisions would be energetic enough to react at (a) $100 .{ }^{\circ} \mathrm{C}$ (b) $200 \cdot{ }^{\circ} \mathrm{C}$ (c) $500 .{ }^{\circ} \mathrm{C}$ (d) $1000 .{ }^{\circ} \mathrm{C}$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD