Use your answer from Problem 7.54 to calculate the following:
(a) \(\left[\mathrm{O}_{2}\right]\) at equilibrium when \(\left[\mathrm{CO}_{2}\right]=0.18 \mathrm{~mol} / \mathrm{L}\) and [CO]=0.0200 mol/L
(b) \(\left[\mathrm{CO}_{2}\right]\) at equilibrium when [CO]=0.080 mol/L and \(\left[\mathrm{O}_{2}\right]=0.520 \mathrm{~mol} / \mathrm{L}\)
Text Transcription:
[O_2]
[CO_2]=0.18 mol/L
[CO_2]
[O_2]=0.520 mol/ L
Chapter 5: Chemical bonding I 1. Bond characteristics - Pure (nonpolar bond) covalent bond: electrons shared equally - Polar covalent bond: electrons shared unequally - Ionic Bond: electrons transferred 2. VSEPR - Basis: pairs of valence electrons in bonded atoms repel one another 3. States of matter - Gas molecules move with great energy >>> overcome intermolecular force >> intermolecular force of little importance 1. Intermolecular forces - Determine melting – freezing points and other physical properties - Types of intermolecular forces: + Dispersion/London Force + Dipole – Dipole + H-Bonding + Ion – Dipole - More electrons >> higher molecular weight >>> Stronger interactions >>>> Higher boiling points. a. Dispersion - Attraction between positive + and negative – ends of nonpermanent dipoles - All molecules have Dispersion forces - For same number of e/molecular weight >>>polar molecules have higher boiling point than nonpolar ones <<<< stronger intermolecular forces in Polar molecules b. Dipole – Dipole - Dipole – Dipole >> Dispersion Because Dipole – Dipole has permanent dipoles - Polar molecules have permanent dipoles - Dipole – Dipole only exists in Polar molecules, not in Nonpolar. c. Hydrogen bond - Hydrogen bond is a type of Dipole – Dipole bond - Hydrogen bonds need: + Partially positive H atom + lone e pair on small electronegative ato
