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Surface Chemistry Test - 52

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Surface Chemistry Test - 52
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  • Question 1
    1 / -0
    Consider the following sequence of reactions:

    Red Hot Coke$$\ +\ $$Steam $$\xrightarrow {\displaystyle \text{Y}{(\text{catalyst})}} X \longrightarrow Z  \xrightarrow [\displaystyle \text{cold water}]{\displaystyle \text{pressure}}\ $$W(gas)

    $$\text{Y}$$ is:
  • Question 2
    1 / -0
    Select the wrong statement.
    Solution
    $$\begin{array}{l}\text{ The decomposition of }\mathrm{H}_{2}\mathrm{O}_{2}(\mathrm{l})\text{ is heterogenous }\text{catalysis because }\mathrm{O}_{2}(\mathrm{~g})\text{ is formed in }\\\text{ the chemical reaction which is in gaseous }\text{ state .}\end{array}$$
  • Question 3
    1 / -0

    Directions For Questions

    The catalytic activity and collodial nature of a substance are surface phenomenon. Both these properties depend upon the property of adsorption. Adsorption may be physisorption or chemisorption. Adsorption is spontaneous and always leads to decrease in entropy and evolution of heat. Chemisorption is irreversible with temperature, unilayer, specific and directional. Adsorbate molecules adsorbs on cataylst surface and thus, lower energy of activation of reaction to provide a new path way for reaction. In colloidal state, dispersed phase particles possess the adsorption characteristics at the interface.

    ...view full instructions

    Which statements are correct?
    1. Addition of $$AgNO_3$$ drop by drop in excess of $$KI$$ leads to the formation of $$+$$vely charged sol of $$AgI$$.
    2. Adsorption characteristic of an adsorbent may be increased by making it in finely divided form.
    3. Catalyst help in attaining the equilibrium earlier but does not influence equilibrium constant in reversible reactions.
    4. Decolorisation of raw sugar is based on absorption.
    5. Gas masks containing animal charcoal acts on the basis of selective adsorption.
    6. The micelle formation of soaps depends upon hydrophobic and hydrophilic entities present in soap.
    Solution
    1. Addition of $$AgNO_3$$ in $$KI\:(excess)$$ gives $$-$$ve sol due to adsorption of $$I^-$$ on $$AgI$$, i.e., preferential adsorption.

    2. An increase in surface area of adsorbent increases the adsorption rate.

    3. The forward reaction is catalysed and takes place fastly whereas the increase in products concentration makes reverse reaction fast and thus, equilibrium is attained earlier.

    4. Animal charcoal adsorbs colouring matter of sugar.

    5. $$CH_4$$ and $$CO$$ are preferentially adsorbed on animal charcoal

    6. The micelle formation of soaps depends upon hydrophobic and hydrophlilic entities present in soap and it is reason why soaps forms micelles.
  • Question 4
    1 / -0
    Smoke precipitator works on the principle of :
    Solution
    Electrostatic smoke precipitators work by forcing dirty flue gas past two electrodes. The first electrode is charged with a very high negative voltage. As the dirt particles move past it, they pick up a negative charge. 

    There is a second electrode consisting of metal plates charged to a high positive voltage. Since unlike charges attract, the negatively charged particles are attracted to the positively charged plates and stick there.
  • Question 5
    1 / -0
    Mention catalyst in the following reaction:
    $$CS_2+3Cl_2 \rightarrow  CCl_4 + S_2Cl_2$$ 
    Solution
    Carbon disulfide reacts with chlorine to produce carbon tetrachloride and disulfur dichloride. 
    In this reaction, the catalyst could be 1) Manganese (ll) chloride - $$MnCl_2$$
                                                                 2) Aluminium  Chloride - $$AlCl_3$$
                                                                 3) Iron(lll) Chloride - $$FeCl_3$$
  • Question 6
    1 / -0
    Which of the following is not correct for heterogeneous catalysis?
    Solution
    Catalyst is a substance that lowers the activation energy of reaction because it provides an alternate pathway for reaction. Also, it increases the rate of backward and forward reactions to the same extent. Due to this lower activation energy, equilibrium attains faster but it doesn't change overall enthalpy of reaction.

    In some cases, the catalyst reacts with reactant and at the end of reaction it forms again. 

    Heterogeneous catalysis refers to the form of catalysis where the phase of the catalyst differs from that of the reactants. Phase here refers not only to solid, liquid vs gas but also immiscible liquids.

  • Question 7
    1 / -0
    When dilute aqueous solution of $$AgNO_3$$ (excess) is added to $$KI$$ solution, positively charged sol particles of $$AgI$$ are formed due to adsorption of ion :
    Solution
    Solution particles possess the tendency to adsorb preferentially the common ion present in solution as follows:

    $$AgI+Ag^+ \rightarrow Ag_2I^+$$

    So, option B is correct.
  • Question 8
    1 / -0

    Directions For Questions

    In macromolecular type of colloids, the dispersed particles are themselves large molecules (usually polymers). Since these molecules have dimensions comparable to those of colloidal particles, their dispersions are called macromolecular colloids. Most lyophilic sols belong to this category. There are certain colloids which behave as normal strong electrolytes at low concentrations but exhibit colloidal properties at higher concentrations due to the formation of aggregate particles. These are known as micelles or associated colloids. Surface active agents like soaps and synthetic detergent belong to this class.
    Critical micelle concentration (CMC) is the lowest concentration at which micelle formation appears. CMC increases with the total surfactant concentration. At a concentration higher than CMC, they form extended parallel sheets known as lamellar micelles which resemble biological membranes. With two molecules thick, the individual molecule is perpendicular to the sheets such that hydrophilic groups are on the outside in aqueous solution and on the inside in a non-polar medium.
    In concentrated solution, micelles take the form of long cylinders packed in hexagonal arrays and are called lyotropic mesomorphs.
    In an aqueous solution (polar medium), the polar group points towards the periphery and the hydrophobic hydrocarbon chains point towards the center forming the core of the micelle. Micelles from the ionic surfactants can be formed only above a certain temperature called the Kraft temperature. They are capable of forming ions. Molecules of soaps and detergents consist of lyophilic as well as lyophobic parts which associate together to form micelles. Micelles may contain as many as $$100$$ molecules or more.

    ...view full instructions

    Select incorrect statement(s):
    Solution

    Option (C) is incorrect because $$C_{17}H_{35}$$ is hydrophobic, whereas $$COO^-$$ of stearate ion ($${ C }_{ 17 }{ H }_{ 35 }{ COO }^{ - }$$) is hydrophilic. This owes to the polarity of ions and hydrocarbons.

  • Question 9
    1 / -0
    Select the correct statements about preparation of colloidal solution.
  • Question 10
    1 / -0

    Directions For Questions

    Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the contact process, vanadium compounds in the +5 state ($$V_2O_5$$ or $$VO_3{^-}$$) are used to oxidise $$SO_2$$ to $$SO_3$$ : $$SO_2$$ + $$\displaystyle{\frac{1}{2}}O_2 \overset{V_2O_5} {\rightarrow}$$ $$SO_3$$. It is thought that the actual oxidation process takes place in two stages. In the first step, $$V^{5+}$$ in the presence of oxide ions converts $$SO_2$$ to $$SO_3$$. At the same time, $$V^{5+}$$ is reduced to $$V^{4+}$$. $$2V^{5+} + O^{2-} + SO_2 \rightarrow 2V^{4+} + SO_3$$. In the second step, $$V^{5+}$$ is regenerated from $$V^{4+}$$ by oxygen : $$2V^{4+} + \displaystyle{\frac{1}{2}O_2 \rightarrow} 2V^{5+} + O^{2-}$$. The overall process is, of course, the sum of these two steps: $$SO_2 + \displaystyle{\frac{1}{2}}O_2 \rightarrow SO_3$$.

    ...view full instructions

    During the course of the reaction:
    Solution
    Vanadium  goes from$${ V }^{ 5+ }\rightleftarrows { V }^{ 4+ }$$ so there is change in oxidation state
    using $$K=A{ e }^{ \cfrac { -{ E }_{ a } }{ RT }  }$$ a catalyst lowers $${ E }_{ a }$$ thus increasing K
    A catalyst like$${ V }^{ 5+ }$$ here is regenerated by$$\cfrac { 1 }{ 2 } { O }_{ 2 }$$ as shown in the questions
     thus all are correct
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