Part III: Examining Changes in Terms of Energy in .NET framework

Maker European Article Number 13 in .NET framework Part III: Examining Changes in Terms of Energy

Consider the following reaction, at equilibrium: 2SO2(g) + O2(g) 2SO3(g) In which direction will each of the following perturbations shift the system a Adding SO3 b Removing O2 c Pumping unreactive neon gas into the reaction vessel d Introducing a metallic catalyst into the system

a Adding SO3 pushes the reaction to the left b Removing O2 pulls the reaction to the left c Pumping unreactive gas into the reaction vessel increases the pressure The system shifts in the direction that reduces the total moles of gas This direction is to the right, because the product state contains only 2 moles of gas for every 3 moles of gas in the reactant state d Introducing a catalyst produces no response, because catalysts alter only the kinetics of a chemical reaction, not its equilibrium (Yes, this part was a trick question!)

Octane, C8H18, is a major component of gasoline Octane vapor combusts to yield carbon dioxide and water vapor: 2C8H18(g) + 25O2(g) 16CO2(g) + 18H2O(g) + heat a How does increasing the partial pressure of oxygen affect combustion b How does increasing temperature affect combustion c Within engines, fuel and air are compressed within a cylinder and then ignited The combusting fuel-air mixture expands, driving a piston through the cylinder How does the expansion affect combustion

You conduct several trials of a mysterious, gas phase reaction The reaction includes an unknown reactant, X, and unknown product, Y: nX + 2O2 4Y You observe that the equilibrium yield of Y at 298K is smaller than the yield of Y at 273K Furthermore, you observe that the yield of Y at 3 atm is smaller than the yield of Y at 1 atm a Does the reaction absorb heat (endothermic) or give off heat (exothermic) (See 15 for more information about these terms) b Which of the following values of n is most plausible: 1, 2, or 3

You may have raced through these problems, or you may have moved at the speed of a dead snail in winter But rate is separate from equilibrium whatever your pace, you ve made it this far Now shift in opposition to any perturbing problems; check your work

The reaction equation makes clear that each mole of methane reactant corresponds to 1 mole of carbon dioxide product and 2 moles of water product So: d[CH4] / dt = d[CO2] / dt = 05d[H2O] / dt This means that the disappearance of 279 mol s 1 of methane corresponds to the appearance of 279 mol s 1 of CO2 and the appearance of 2 279 = 558 mol s 1 of H2O The concentration of methane changes at half the rate of water, so the concentration of water changes at twice the rate of methane

The observations are consistent with the rate law, Rate = k[A]2[B] Tripling [A] increased the rate ninefold, and 32 = 9 Doubling [B] doubled the rate, and 21 = 2 The overall reaction order (3) is the sum of the individual reaction orders (the exponents on the individual reactant concentrations), 2 + 1 = 3 Doubling the concentration of either reactant (D or E) doubles the rate So, the data are consistent with the rate law, Rate = k[D][E] Solve for k by substituting known values of rate [D] and [E] from any of the trial reactions: k = Rate / ([D][E]) = 48 106M s 1 / (27 10 2M 27 10 2M) = 66 109M 1 s 1 Use this calculated value of k to determine the rate in the presence of 13 10 2M reactant D and 092 10 2M reactant E: Rate = (66 109M 1 s 1)(13 10 2M)(092 10 2M) = 79 105M s 1

The corning process converts gunpowder from a finely divided powder of undefined particle size into larger particles of defined size Because the corned particles are larger, combustion occurs more slowly Corned gunpowder is less likely to explode accidentally than is a fine dust of gunpowder Because the corned particles are all the same size, combustion occurs at a predictable rate, which is convenient for the user The change in conditions does result in a faster rate Rate is limited by activation energy, the difference in energy between reactant (A) and the transition state (A*) If the energy of A is raised and the energy of A* remains constant, then the difference in energy (the activation energy) becomes smaller, and lower activation energies result in faster reactions The decrease in the energy of product B has no bearing on the rate because it has no effect on the activation energy for the A B reaction although it would affect the rate of the reverse reaction, B A Here are the answers regarding equilibrium in the reaction, N2(g) + O2(g) 2NO(g) a The Keq = (263 10 7 atm)2 / [(476 10 2 atm)(982 10 3 atm)] = 148 10 10 b Q = (263 10 7 atm)2 / [(476 10 2 atm)(374 10 2 atm)] = 389 10 11 Because Q < Keq, the reaction would proceed to the right, converting reactant into product