Download Reversible Reactions and more Study notes Chemistry in PDF only on Docsity!
Denoted by The reaction that proceeds both in forward and backward directions under same conditions. Product yield would vary stoichiometrically while the reactants would not be fully used up and will reach dynamic equilibrium. The dotted line shows the time when the reaction reaches dynamic equilibrium. For graph one it is the graph showing that the equilibrium is towards left as the concentration of the reactants are more than the products, usually in the case of weak acids. For graph two, it is shifted towards the right hand side as more products are formed, usually in the case of the strong acids. Irreversible reactions are the reactions that undergoes complete reaction and is always forward. As is a complete reaction so the reactants should decrease to 0. Dynamic equilibrium: it is a state in a reversible reaction when the rate of forward reaction equals t he rate of backward reaction in a closed container. Haber process: Nitrogen react with hydrogen to form ammonia. Iron as a catalyst. 450 degree Celsius temperature needed. Reversible reactions.
200 atm pressure needed. Occurs in a closed container.
Le chatliers principle:
if a system in equilibrium is disturbed by any external factors then the system will adjust itself in a such a way so as to undo or minimize the effect of the change. effect on change in concentration on equilibrium: a) reactants concentration increase by adding reactants so the equilibrium will shift to the right hand side as the reaction will be forward and more products will form. b) Products concentration increase so the equilibrium will shift to the left hand side as backward reaction starts however the yield will decrease. c) Products to be removed from the container so forward reaction will start and equilibrium shifts to the right hand side while the yield will increase. Effect of change in pressure on equilibrium:
- It is applicable only upon gases.
- Increase in pressure will favour the side with fewer moles.
- Decrease in pressure will favour the side with more moles. Effect on change in temperature on equilibrium:
RULE:
If the temperature increases, the system would work in a way so as to
reduce the heat gained hence there would be endothermic reaction. While if
the temperature is decreased so there would be exothermic reaction will be
favoured. remember that in the case of temperature any change would not
establish the equilibrium again.
For example N 2 + H 2 ===NH 3 delta H is -410kJ/mol So as enthalpy is negative hence the forward reaction is exothermic. Case1: if the temperature is increased what would be the effect of this change on the equilibrium? Ans : as the temperature is increasing so there would be more heat hence there should be endothermic reaction hence the backward reaction takes place. Hence the yield would decrease and it will keep on decreasing until the temperature is again decreased. Case2: if the temperature is decreased what would be the effect of it on the equilibrium? Ans: as the temperature decreases so there should be exothermic reaction which is forward hence there would be more yield of ammonia.
N 2 + 3H 2 === 2NH 3 all are in gaseous state. KP = (p NH 3 )^2 /(p N 2 ).(p H 2 )^3 And the unit of it will be if demanded in pascals so becomes Pa^2 /Pa^4 so becomes Pa-2.
Mathematical calculations :
Q N 2 + 3H 2 === 2NH 3
Where when the equilibrium is established N 2 HAS 2 moles while hydrogen has 4 moles and ammonia has 2 moles while the volume of the close vessel is 1 dm^3. Find kc. Ans: [NH 3 ]^2 /[N 2 ]^2 .[H 2 ]^4 while their concentrations would be the same as the moles because the concentration will be moles / volume. 4/1024 mol-2dm^6. Q N 2 O 4 === 2NO 2 In a flask of volume 100 cm^3 , 2.5 moles of N 2 O 4 were placed and sealed and allowed to reach equilibrium. At equilibrium it was found that 1.2 mol of N 2 O 4 were present. Calculate Kc. ANS: initial moles of N2O4=2.5 and final was 1.2. The difference is of 1.3moles. N2O4:NO 1: 1.3:x X=2.6 moles means that 2.6 moles of NO2 gas are present at equilibrium. Now we are to find the concentrations so remember that the basic unit is mol/dm^3. So we have volume 100cm^3 that is equal to 0.1 dm^3. So 1.2/0.1=concentration of N2O4 and 2.6/0.1= concentration of NO Now apply in the equation 2.6^2 /12 moldm-3. Q 2SO 2 + O 2 === 2SO 3 When the 2 mol of SO 2 and 1.2 mol of oxygen gas and 0.8 moles of SO3 were placed in a closed vessel of volume 250cm^3 were allowed to reach equilibrium. It was found that mol of SO 3 were 1.4. Initial mole of SO2 = INITIAL MOLES OF O2=1. SO3 MOLES= 0.8 while final moles are 1.4 so 0.6 are gain.
Now figure out the moles of the reactants using the ratio method of moles. FINAL MOLES OF SO2=2-0.6=1.4 as we are to find the moles at equilibrium so if 0. moles of SO3 are made then 0.6 moles of SO2 would be used up similar with O2. The final moles of O2 will be 1.2 -0.3=0.9. NOW figure out their concentrations and convert 250 cm^3 to dm^3. SO2 has 5.6 , o2 has 3.6 while SO3 has 5.6 concentrations. Now apply the formula and find its Kc. The final answer will be 0.278 mol-1dm^3. Here for HI there were 0.8 moles used up so we would have 0.4 moles left behind
Mathematical calculations of Kp:
Remember that the term partial pressure refers to that pressure possessed by the individual gas. While the term pressure refers to the sum of all partial pressures. In case of the reaction between gases to form a gas, the pressure would be the sum of the partial pressures of the reactants and the products. Partial pressure= mole fraction. Total pressure Where mole fraction= mol of gas at equilibrium/total moles of all gases at equilibrium. Q N 2 + 3H 2 == 2NH 3 In an experiment 0.2 moles of nitrogen and 0.5 moles of hydrogen were placed in a closed vessel and allowed to reach equilibrium and at equilibrium there were 0. moles of nitrogen. The pressure of the vessel was 50Kpa calculate Kp. ANS: firstly calculate the moles of hydrogen and ammonia at the equilibrium by ratio method using the moles of nitrogen. The moles of hydrogen used up should be subtracted from the initial moles as we are to find the remaining moles. 0.3 moles are used up and we are left with 0.2 moles of hydrogen. While ammonia was not present
