Surface Chemistry Solved Problems

SURFACE CHEMISTRY SOLVED PROBLEMS Q.1 Ans. Give the sign of H and S for adsorption process. Or What is the sign of H and S when bromine gas gets adsorbed on charcoal? H is –ve (enthalpy decreases) and S = – ve (entropy decreases). Q.2 Ans. How can we make dialysis fast ? By applying electric field. Q.3 Ans. Write down the heterogeneous catalyst involved in the polymerisation of ethylene. ((C2H5)3Al + TiCl4) Zeigler – Natta catalyst. Q.4 Ans. Artificial rain can be caused by spraying common salt on the clouds. How ? Clouds are colloidal dispersion of water particles in air. These colloidal water particles cause some change on applying salt in the cloud, coagulation of water particles takes place. Q.5 Ans. What do you understand by activation of adsorbent? How is the achieved ? Activation of adsorbent : The phenomenon which involves the increasing of the adsorbing power of an adsorbent by suitable means is known as activation of adsorbent. Activation of adsorbent refers to its clean surface and fine particle size. Cleaning can be done by ultra high vacuum while large surface area can be achieved by grinding. Q.6 Ans. What role does adsorption play in heterogeneous catalysis? Heterogeneous catalysis generally proceeds through the adsorption of reactants on the surface of catalyst. In such type of catalysis generally adsorbent is solid while the reactants are gaseous. The reactions occur at the surface of the catalyst where the gaseous molecules get adsorbed either by physical adsorption or by chemical adsorption. Q.7 Ans. Why is adsorption always exothermic? When a gas is adsorbed on the surface of solid its entropy decreases, it means s is –(ve). Now, G = H – TS and for the process to be spontaneous, free energy change (G),must be negative. If H has sufficient high –(ve) value as TS is positive. Thus, in an adsorption process, which is spontaneous, a combination of these two factors makes G negative. As the adsorption proceeds, H become less and less negative ultimately H becomes equal to TS and G become zero at equilibrium state. Q.8 Ans. Action of soap is due to emulsification and micelle formation. Comment. Soap molecules adsorb on to the dirty (oily) surface and try to emulsify it. Soap micelles do solubilize the insoluble dirt as greasy material. Q.9 Ans. What do you mean by activity and selectivity of catalysts? Activity : Activity of a catalyst refers to the chemisorption of reactants reasonable strongly. It has ability to increase chemical reaction. For example : Transition metals show catalytic activity. (Maximum activity group 7-9 metals). Selectivity : Selectivity of a catalysts is its ability to speed up a reaction to yield a particular product. For example: CO + H2 combine to form CH4 when Ni is the catalyst and methanal when Cr2O3 is catalyst. CO + 3H2 CO + 2H2 CH4 + H2O Pt Cu/ZnO-Cr2O3 CO + H2 Cu CH3OH HCHO Methane Methyl alcohol Formaldehyde SURFACE CHEMISTRY Q.10 Explain the term with suitable examples : (i) Alcosol Ans. (i) (ii) Aerosol Alcosol : (iii) Hydrosol. Alcosols are those sols in which the dispersion medium is alcohol. For example : Collodion (a colloidal solution of cellulose nitrate in ethyl alcohol). (ii) Aerosol : Aerosol are those sols in which the dispersion medium is air. For example : Smoke, dust. (iii) Hydrosol : Hydrosol are those sols in which the dispersion medium is water. For example : Starch solution, gold solution. Q.11 Comment on the statement that "colloid is not a substance but a state of substance." Ans. Colloidal solution is made up of dispersed phase and dispersion medium. The size of the particles of dispersed phase varies between 1 nm to 1000 nm in colloidal solution. On the basis of size of particles, we can distinguish between colloidal solution and true solution. For example : NaCl in water forms true solution but in benzene forms colloidal solution. A soap forms true solution in alcohol but colloidal solution in water. Q.12 Discuss the origin of charge on colloidal particles. Or How does the theory of selective adsorption, explain the origin of charge on colloidal particles. Ans. Due to selective adsorption of ions: These particles constituting the dispersed phase adsorb those ions preferentially which are common with their own lattice ions. For example, if silver nitrate solution is added to an aqueous solution of potassium iodide, the silver iodide will adsorb the I– (Common ions) from the dispersion medium to form a negatively charged sol. + I– Ag + + Ag + – Ag I Ions Absorbed + Ag Ag Ag + Ag + + + + AgNO3 + KI  AgI AgNO3 Ag Ag ions remain in dispersion medium  Dispersion AgI/I– Negative sol. medium However, if silver iodide is formed by adding potassium iodide to silver nitrate solution, the sol is positively charged due to the adsorption of Ag+ ions (Common ions) present in the dispersion medium. I I I AgI AgI Ag+ + —  Dispersion medium AgI / Ag+ Positive sol. I — + — — I I— KI + AgNO3  — — I I Ag Ions Absorbed — — I ions remain in dispersion medium