Quiz # 2 SHOW ALL WORK FOR CREDIT R2.0 cal/mole OK the rolowing re aodiong whicf
ID: 704354 • Letter: Q
Question
Quiz # 2 SHOW ALL WORK FOR CREDIT R2.0 cal/mole OK the rolowing re aodiong whicf vill ncrceas the equllbium amount of S03 in 2 SO2(g) + O2(9)-2503(g) ?»--57 Kcal 1. Circle all of the the following reaction: w a. Removing SO2 b: Adding 02 c. Raising the pressure (dec. V) d. Adding He e. Raising the temperature 2. (a) Write Kp for the equation in question #1: (b) For which answers in question #1 does KC change? 3. Calculate Kc at 298 oK if, at eauilibrium, we find a 2.0 liter flask contains 0.020 moles of PCls, 4.0 moles of PCI3 and 10.0 moles of Cl2: PCI3(9) Cl2(9)PCIs(g) Calculate ?G298 ifK_ 0.00010 4. (OVER)Explanation / Answer
1) The given equilibrium reaction is
2 SO2 (g) + O2 (g) <======> 2 SO3 (g) ?H = -57 Kcal
Write down the equilibrium constant for the reaction as
KC = [SO3]2/[SO2]2[O2] ……(I)
a) When SO2 is removed from the system, KC tends to increase. However, KC is a thermodynamic equilibrium constant and hence, must remain constant at a particular temperature. To keep KC constant, [SO3] must decrease, i.e, the reaction proceeds to the left.
b) When O2 is removed from the system, [O2] falls and KC tends to increase. However, the equilibrium reaction adjusts itself in such a way as to keep KC constant. This is possible only when the concentration of SO3 falls, i.e, the reaction again moves to the left.
c) There are 3 moles of gaseous reactants while 2 moles of gaseous products. Thus, the reaction proceeds with a change in the number of moles of gases by (2 – 3) = -1, i.e, ?n = -1. As per the gas laws, we have P*V = n*R*T where P, V and T denote the pressure, volume and temperature of the gaseous system. It is clear that P is directly proportional to n. When the pressure of the system is increased, the system will adjust itself in such a way that the effect of increased pressure is nullified, i.e, the reaction will move in the direction where the number of moles of gases decreases. Since, there is a decrease in the number of moles on the product side, hence, increasing the pressure will favor the forward reaction, i.e, the concentration and hence, amount of SO3 at equilibrium will increase.
d) He doesn’t take part in the reaction. Hence, if the equilibrium reaction takes place in a container of fixed volume, adding He will have no effect on the concentration of the reactants and the products at equilibrium
e) The reaction releases heat, i.e, the reaction is exothermic. Exothermic reactions are favored by a lowering of temperature. Infact, the value of KC decreases at high temperatures and thus, the concentration of the product must decrease.
2a) Since the system comprises of gaseous reactants and products, hence, the equilibrium constant (KP) can be written as
KP = (PSO3)2/(PSO2)2(PO2) (ans).
b) The value of KC will change only when there is a change in temperature of the system. Hence, (e) is the only option which can change the value of KC.
3) The equilibrium reaction is given as
PCl3 (g) + Cl2 (g) ---------> PCl5 (g)
The equilibrium constant for the reaction KC is given as
KC = [PCl5]/[PCl3][Cl2]
where the square braces denote molar concentrations.
We can find out the molar concentration of the reactants and the products simply by dividing the number of moles of each gas at equilibrium by the volume of the system, i.e,
[PCl3] = (number of moles of PCl3)/(volume of the flask) = (4.0 moles)/(2.0 L) = 2.0 mol/L
[Cl2] = (number of moles of Cl2)/(volume of the flask) = (10.0 moles)/(2.0 L) = 5.0 mol/L
[PCl5] = (number of moles of PCl5)/(volume of the flask) = (0.020 mole)/(2.0 L) = 0.010 mol/L.
Therefore,
KC = [PCl5]/[PCl3][Cl2] = (0.010 mol/L)/(2.0 mol/L)(5.0 mol/L) = 0.0010 L/mol.
KC is usually expressed as a dimensionless quantity; hence, the value of KC is 0.0010 (ans).
4) We know that
?G0 = - R*T*ln K
where T is the absolute temperature of the system. Given K = 0.0010 and T = 298 K, we have,
?G0 = -(2 cal/mol.K)*(298 K)*ln (0.0010)
= -(596 cal/mol)*(-6.9077)
= 4116.9892 cal/mol
= (4116.9892 cal/mol)*(1 Kcal/1000 cal) =4.1169892 Kcal/mol ? 4.12 Kcal/mol (ans).