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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Hydrogen peroxide decomposes to water abd oxygen. The uncatalysed reaction has activation energy of 86 KJ/mol. The activation energy value in the presence of acetanilide is 112 KJ/mol and in the presence of `MnO_(2)` it is 49 KJ/mol . What conclusion can you draw from the above observations ? |
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Answer» The activation energy of a catalysed reaction in presence of `MnO_(2)` catalyst is 49 KJ/mol that is much less than the activation energy of an uncatalysed reaction that is 86 KJ/mol . This show that the rate of decomposition of `H_(2)O_(2)` is increased in the presence of `MnO_(2)` and hence it is a positive catalyst. THe activation energy of the catalysed reaction in the presence of acetanilide is 112 KJ/mol that is such greater than that of an uncatalysed reaction. This shows that the rate of decomposition of `H_(2)O_(2)` is decreased in the presence of acetanilide and hence it is a negative catalyst. |
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| 2. |
`K_(c)` changes with change in _______. |
| Answer» Correct Answer - temperature | |
| 3. |
The decomposition reaction `N_(2)O to N_(2) + O_(2)` takes place on platinum surface. Here, the rate of reaction is independent of the concentrations of the reactant . However, when this reaction is carried out in the absence of platinum surface, the rate of reaction depends on the concentration of the reactant How do you account for this ? |
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Answer» (i) role of plantinum in the reaction (ii) property of platinum (iii) site of decomposition (iv) comparison of mechanism of reaction in the presence and absence of a catalyst |
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| 4. |
If `K_(C)` for the formatiion of ammonia is 2 `mol^(-2) L^(2)`, `K_(c)` for decoposition of ammonia is ________. |
| Answer» Correct Answer - `0.5 mol^(2)//1^(2)` | |
| 5. |
The equilibrium constant for the given reaction, ` CaCO_(3(g)) rarr CaO_(s) +CO_(2(g)) ` is given by :A. `K_(c)=([CaO]*[CO_(2)])/([CaCo_(3)])`B. `K_(c)=([CaO])/([CaCO_(3)])`C. `K_(c)=[CO_(2)]`D. `K_(c)=([CaO])/([CO_(2)])` |
| Answer» Correct Answer - C | |
| 6. |
Which of the following `k_(c)` values corresponds to the maximum yield of the product ?A. `9.2 xx 10^(2)`B. `1.8 xx 10^(-15)`C. `2.8 xx 10^(3)`D. `3.4 xx 10^(-25)` |
| Answer» The greater the `K_(c)` value, greater is the yield of products. | |
| 7. |
A 1 L reaction vessel contained 1 mole each of solid `NH_(4)HS, NH_(30 and H_(2)S` at a temperature of `150^(@)C` . When the decomposition of `NH_(4)HS` was carried out , equilibrium is established `K_(p)` value at that temperature is 100. Calculate the equilibrium partial pressures at which 60 per cent dissociation of `NH_(4)HS` takes place at a lower temperature where `K_(p)` value is equal to 200 `"atm"^(2)` . |
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Answer» `NH_(4)HS_((s)) hArr NH_(3(g)) + H_(2)S_((g))` `1-x" "1+x" "1+x` `K_(p)=P_(NH_(3)) xx P_(H_(2)S)` Total no. of moles at equilibrium `=(1+x)+(1+x)=2x+2` Partial pressure=Mole fraction `xx`P Mole fraction of `NH_(3)=(1+x)/(2x+2)` `therefore ` Partial pressure of `NH_(3)=(1+x)/(2x+2)xxP` Mole fraction of `H_(2)S=(1+x)/(2x+2)` `K_(p)=((1+x)/(2x+x)xxP)((1+x)/(2x+2)xxP)` `K_(p)=((1+x)^(2))/((2x+2)^(2))=100` `((1+x)^(2)P^(2))/(4(1+x)^(2))=100 implies P^(2)=100xx4` `P=sqrt(400)=200 ` atm `P_(H_(2)S)=((1+x)20)/(2(x+1))=10`atm `P_(NH_(3))=10` atm per cent dissociation of `NH_(3) and H_(2)S` corresponds to moles each of `NH_(3) and H_(2)S` `K_(p)=P_(NH_(3))xxP_(H_(2)S)` `P_(NH_(3))=(1.6)/(3.2)xxp,P_(H_(2)S)=(1.6)/(3.2)xxP` `K_(P)=((1.6)xxP^(2))/((3.2)^(2))=400` `P^(2)=(200xx(3.2)^(2))/((1.6)^(2))=400` `P=sqrt(400)=20` atm |
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| 8. |
Assuming that `2NO + O_(2) to 2NO_(2)` is a single-step reaction, what will be the rate of reaction when the volume of the reaction vessel is reduced of `1//4^(th)` of the initial value ? The original rate of reaction is 64 mol/L/s. |
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Answer» `2NO + O_(2)to 2NO_(2)` The rate of the reactioin r`=k[NO]^(2)[O_(2)]`, as it is a single-step reaction. When the volume of the reaction vessel is reduced to `1//4^(th)` of the initial volume, the concentration of each reactant is increased by four-times. Since the reaction is a second order with respect to NO and first order with respect to `O_(2)` , the rate of reaction increases by 16-times for NO and 4-times for `O_(2)`. On the whole , the rate of reaction increases by 64-times . The original rate of reaction is 64 mol/L/s and the new rate of reaction is 64-times more than this. `r_(2)=64xx64=4096=4.096xx10^(3)` mol/L/s. |
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| 9. |
Identify the common property for a chemical reaction at dynamic equilibriumA. The measurable properties like concentration , density , colour, pressure , etc. , remain constant at constant temperature .B. The forward and backward reactions take place with the same rate.C. It can be achieved from both directionsD. all of the above |
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Answer» (1) The measurable properties like concentration density, colour , pressure etc., remain constant at constant temperature . (2) The forward and backward reactions take place with same rate. (3) It can be achieved from either direction. |
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| 10. |
`2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) + Q ` KJ In the above reaction, how can the yield of product be increased without increasing the pressure ?A. by increasing temperatureB. by decreasing temperatureC. by increasing th volume of the reaction vesselD. by the addition of the catalyst |
| Answer» Correct Answer - B | |
| 11. |
Initial number of moles of reactants taken in a closed reaction vessel is given. Percentage degree of dissociation is also given . Identify the correct sequence of steps to calculate `k_(c)` value. (1) calculation of equilibrium concentrations of reactants and products (2) calculation of equilibrim concentration number of moles (3) writing equilibrium constant expression for the reaction (4) calculation of `k_(c)` value by using the equilibrium concentrationA. 4,2,3,1B. 2,1,3,4C. 3,2,1,4D. 2,1,4,3 |
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Answer» (i) calculate of equilibrium number of moles (ii) calculation of equilibrium concentrations of reactants and products (iii) writing equilibrium expression for the reaction (iv) calculate of `k_(c)` value by using the equilibrium concentration |
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| 12. |
In the reaction `N_(2) + O_(2) hArr 2NO-` Heat , which of following conditions is suitable to get a good yield of NO ?A. Increase in temperatureB. Decrease in temperatureC. Increase in pressureD. The addition of a catalyst |
| Answer» Correct Answer - A | |
| 13. |
During the preparation of soap the addition of common salt allows the precipitation of soap . Expalin the principle involved (Soap is the sodium salt of carboxylic acid ). |
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Answer» The addition of comon salt results in the addition of `Na^(+)` ions. Equilibrium exist between sodium and stearate ions. The addition of `Na^(+)` ions shifts the equilibrium towards left thus sodium stearte precipitates. `RCOO^(-)Na^(+) hArr RCOO^(-)+Na^(+)`. |
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| 14. |
For a reaction `2A + B to 2AB` , it is found that doubling the concentration of both the reactants increases the rate to eight times that of initial rate but doubling the concentration of B alone doubles the rate. Then the order of the reaction with respect to A and B isA. 0,3B. 0,2C. 2,1D. 2,2 |
| Answer» Correct Answer - C | |
| 15. |
Under what conditions , addition of insert gas affects the equilibrium position in case of the following equilibrium at constant temperature . Give a reason in support of your answer. (a) decomposition of `NO` to `N_(2) and O_(2)` (b) decomposition of `SO_(3)` (c) formation of `CH_(3)OH ` from `CO and H_(2)` |
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Answer» `2NO hArrN_(2)+O_(2)` The above equilibirum is associated with no change in volume. Therefore additon of inert gas has no effect on the equilibrium under any condiontion. (b) `2SO_(3)hArr2SO_(2)+O_(2)` The above equilibrium is associated with increase in volume. Addition of an inert gas has no effect has no effect at constant volume. At constant pressure, addition of inert gas favours the forward reaction since increases in volume takes place. (c) `CO_((g))+2H_(2(g))hArrCH_(3)OH_((g))` The above equilibrium is associated with decreases in volume. Addtion of the inter gas favours back-ward reaction since increases in volume takes place. |
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| 16. |
Change in temperature results in change in equilibrium position . However, the addition of a catalyst results in no change in the equilibrium position . Justify. |
| Answer» When the temperature is changed, the rate of forward reaction and backward reaction is exothermic, increase in temperature decreases the rate of increase in temperature decreases the rate of forward reaction thereby decreases the concentration of products . If the forward reaction is endothermic , an increase in temperature favours forward reaction and increases the concentration of products . Therefore, the conposition of reaction mixture at equilibrium changes When a catalyst is added, the rate of forward reaction increases due to decrease in activation energy of the forward reaction. Since the reaction takes an alternative path, the activation energy of backward also is reduced . Activation complex is formed at a lower energy state in a catalysed reaction. As both the rate of forward reaction and that of the backward reaction change ot the same extent, the composition of equilibrium mixture remains constant. Therefore, the equilibrium position remains for both catalysed and uncatalysed reactions. | |
| 17. |
Identify the correct sequence of steps in an experiment to show the effect of temperature on the rate of the reaction.(1) Measuring the volumes of `H_(2)` gas liberated in the two test tubes. (2) Heating the test tube B by `10^(@)C` (3) Comparison of relative volumes of `H_(2)` liberated in test tubes B and A. (4) Addition of same concentration of HCl ot the two test tubes. (5) Taking equal masses of fine granules of zinc in two test tubes A and B.A. 3,4,5,1,2B. 5,4,2,1,3C. 2,1,3,5,4D. 5,4,2,3,1 |
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Answer» (i) raking equal masses of fine granules of zinc in two test tubes A and B. (ii) addition of same concentration of HCl ot the two test tubes. (iii) heating the test tube B by `10^(@)C` (iv) measuring the volumes of `H_(2)` gas liberated in two test tubes. (v) observation of volumes of `H_(2)` liberated in test tubes B and A. |
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| 18. |
Equlibrium position of which of the following reactions is not affected by change in pressure ?A. `I_(2(s)) + 5F_(2(g)) to 2IF_(5(g))`B. `FeO_((s)) + CO_((g)) to Fe_((s)) + CO_((g))`C. `2Cu(NO_(3))_(2(s)) to 2CuO_((s)) + 4NO_(2(g)) + O_(2((g))`D. `N_(2)O_(4(g)) to 2NO_(2(g))` |
| Answer» Correct Answer - B | |
| 19. |
If an activated complex is formed in chemical reactions according to the collision theory, which of the following is true with respect to its stability ?A. It is highly stable because it has high energyB. it is less stable because it has lower enerygC. It is less stable because it has high energyD. None of the above |
| Answer» Correct Answer - C | |
| 20. |
(I) What is the effect of pressure on the equilibrium of the reaction between nitrogen and oxygen to give nitric oxide ? (II) In a reversible reaction, some amount of heat energy is liberated in th forward reaction. Name the reaction . What change in temperature favours the forward reaction ? |
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Answer» (i) `N_(2) + O_(2) hArr 2NO` As the total number of moles on reactants side and products side is same, there is no effect of pressure on the equilibrium . (ii) In a reversible reaction, evolution of heat indicates that it is an exothermic reaction and `DeltaH=-ve` For example ,`A + B hArr C+B + ` Energy Hence, low temperature favours the formation of products. |
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| 21. |
In which among the following reactions, the formation of product is favoured by decreasing the temperature or volume ?A. `2SO_(3(g)) hArr 2SO_(2(g)) + O_(2(g)) -q`B. `N_(2(g)) + O_(2(g)) hArr 2NO_(g) -q`C. `4NH_(3(g)) + 5O_(2(g)) hArr 2NO_((g)) + 6H_(2)O_(g) , DeltaH=-ve`D. `2NO_((g)) + O_(2(g)) hArr 2NO_(2(g)), DeltaH=-ve` |
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Answer» In the reaction, `2NO + O_(2) hArr 2NO_(2) + ` Heat Formation of product is favoured by decreasing the temperature or volume. |
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| 22. |
Calculate `K_(p)` for the following reaction if partial pressures of `NH_(3), N_(2) and H_(2)` are 0.4,0.3,0.2, atm, respectively. `2NH_(3) hArr N_(2)+3H_(2)` |
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Answer» `K_(p)=(p^(3)H_(2).pH_(2))/(P^(2)NH_(2))` `K_(p)=((0.2)^(3)*(0.3)("atm")^(4))/((0.4)^(2)("atm")^(2))` `K_(p)=0.015 "atm"^(2)` |
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| 23. |
Why is the reaction of `SO_(2)` to `SO_(3)` not rapid ini clean and dry air ? |
| Answer» The conversion or oxidation of `SO_(2)` to `SO_(3)` takes place in the presence of a catalyst . Water droplets associated with dust particles act as the catalyst for this reaction. The dust particles can absorb the `SO_(2)` on their surface where it can be oxidised to `SO_(3)` . Hence oxidation of `SO_(2)` to `SO_(3)` is not rapid in clean and dry air. | |
| 24. |
An absent-mined prodessie, Mr Waage, took elements A and B in a reaction vessel at room temperature, to study the reaction`A+ 2B rArr 2C +D.` He took the concentration of B as 1.5 times the concentration of A. After the reaction reached equilbrium, he round that the concentrations of A and D were equal. However, he forgot to calculate `K_(c)` and removed oneof the products from the mixture. Now, can you calculate `K_(c)` for the equilibrium attained in his experiment and help him out ? |
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Answer» `A + 2B hArr 2C +D` Initial concentration a 1.5a 0 0 Equilibrium concentration a-x 1.5a-2x 2x x At equilibrium , the concentration of A and that of D are equal `therefore a-x = x implies a=2x ` or ` x=a//2` `K_(c)=((2x)^(2)x)/((a-x)(15a-2x)^(2))` as a=2x `implies (4x^(2).x)/(x xx (x)^(2))implies K_(c) = 4` |
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| 25. |
When two moles of hydrogen are heated with two moels of iodine, 2.96 moles of hydrogen iodide are formed. Calculate `K_(c)` for the reaction of formation of hydrogen iodide. |
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Answer» `H_(2)(g)+I_(2)(g) hArr 2HI(g)` At equilibrium, `2-x" "2-x" "2x` `2x=2.96` `x=(2.96)/(2)=1.48` `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((2x)^(2))/((2-1.48)(2-1.48))=(4xx(1.48)^(2))/(0.52xx0.52)=32.4` |
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| 26. |
A mixture of `0.75` mol of `N_(2) and 1.20` mol of `H_(2)` is placed in a 3 L container. When the reaction `N_(2)+3H_(2) hArr2NH_(3)` reaches the equilibrium, the concentration of `H_(2)` is 0.1 M. Calculate the concentration of `N_(2) and NH_(3)` when the reaction is carried out with double the number of moles. |
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Answer» `N_(2) + 3H_(2) hArr 2NH_(3)` Concentration of `N_(2)=0.75//3=0.25` mol `L^(-1)` Concentration of `H_(2) =1.20//3=0.4 ` mol `L^(-1)` `N_(2) + 3H_(2) hArr 2NH_(3)` `{:("Initial concentration ",0.25,0.4,0),("At equilibrium",0.25-x,0.4-3x,2x):}` Concentration of `H_(2)` at equilibrium is given as =0.1 ` therefore 0.4-3x=0.1 implies x=0.1` Concenration of `N_(2)` at equilibrium =0.25-0.1 =0.15 Concentration of `NH_(3)` at equilibrium =`0.1xx2=0.2` When the reactiion is carried with double the number of moles , concentration of `N_(2)=0.15xx2=0.3` mol `L^(-1)` and concentration of `NH_(3)=0.2xx2=0.4` mol `L^(-1)`. |
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| 27. |
For a gas phase reaction `Cl_(2) + CHCl_(3) to HCl+ "CC"l_(4)`, the rate law is given as `r=k[Cl_(2)]^(1//2) [CHCl_(3)]` . Explain how the rate of reaction varies when the concentration of chlorine is doubled . Give units of rate constant |
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Answer» (i) significant of rate law (ii) equation for rate constant (iii) equation for rate constant with double the concentration of `Cl_(2)` (iv) comparison of both the equations |
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| 28. |
In the reactio : `NO_(2) + CO hArr NO + CO_(2)` , the equilibrium state may be recognised by the constancy of colour. |
| Answer» Correct Answer - 1 | |
| 29. |
What is the use of catalytic convertic in automobile exhaust systems ? |
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Answer» (i) gases produced during comubstion of fuels (ii) nature of gases that are liberated through an exhaust system. (iii) role of catalytic converter for changing the molecules compositioin of the exhaust gases (iv) change in nature of the gases due to the change in the molecules composition |
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