in the Contact Process, sulfur dioxide is converted into sulfur trioxide and the following equilibrium is set up: SO2 (g) + O2 (g)--> 2 SO3 (g) ∆H = -27 kJ a) If sulfur dioxide is added to the mixture, what happens to the position of the equilibrium? b) If a catalyst is added to the mixture, what happens to the position of the equilibrium? c) if the pressure was increased in the above reaction, what happens to the position of the equilibrium? *

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

a is the correct answer