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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.
| 2901. |
Which of the following is an endothermic reaction?A. `2H_(2)(g)+O_(2)(g)rarr 2H_(2)O(l)`B. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`C. `NaOH(aq.)+HCl(aq.)rarr NaCl(aq)+H_(2)O(l)`D. `C_(2)H_(5)OH(aq)+3O_(2)(g)rarr 2CO_(2)+2H_(2)O(l)` |
| Answer» Correct Answer - B | |
| 2902. |
Heat evolved in the reaction `H_(2)+Cl_(2) rarr 2HCl` is 182 kJ Bond energies H- H = 430 kJ/mole, `Cl-Cl = 242kJ//"mole"`. The H-Cl bond energy isA. 763 kJ/moleB. 427 kJ/moleC. 336 kJ/moleD. 154 kJ/mole |
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Answer» Correct Answer - B `Delta H = H_(r) - H_(P)` |
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| 2903. |
The enthalpy change under standard condition for which of the reactions below would be equal to the `DeltaH_(f)^(@)` of NaOH(s)?A. `Na(s)+H_(2)O(l)to NaOH(s)+(1)/(2)H_(2)(g)`B. `Na(s)+(1)/(2)O_(2)(g)+(1)/(2)H_(2)(g)to NaOH(s)`C. `Na(s)+(1)/(2)H_(2)O_(2)(l)to NaOH(s)`D. `Na^(+)(aq)+OH^(-)(aq)toNaOH(s)` |
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Answer» Correct Answer - b |
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| 2904. |
Standard Gibbs energy of reaction `(Delta_(r)G^(@))` at a certain temperature can be completed as `Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@)` and the change in the value of `Delta_(r)H^(@)` and `Delta_(r)S^(@)` for a reaction with temperature can be computed as follows: `Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1))` `Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2)))` `Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) ` `Delta_(r)^(@)G^(@)=-RTInK_(eq)` Consider the following reaction : `CO(g)+2H_(2)(g)toCH_(3)OH(g)` Given `Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol` `Delta_(r)H^(@)(CO,g)=-114KJ//mol`` s^(@)(CH_(3)OH,g)=240J//mol-k,` `S^(@)(H_(2)g)=198J//mol-K``C_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1)``C_(p.m(CO)=29.4J//mol-K` `C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K` `and In ((320)/(300))=0.06,"all data at"300K.` `Delta_(r)s^(@) ` at 320K is:A. 155.18 J/mol-KB. 150.02J/mol-KC. 172J/mol-KD. none of these |
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Answer» Correct Answer - d |
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| 2905. |
Standard Gibbs energy of reaction `(Delta_(r)G^(@))` at a certain temperature can be completed as `Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@)` and the change in the value of `Delta_(r)H^(@)` and `Delta_(r)S^(@)` for a reaction with temperature can be computed as follows: `Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1))` `Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2)))` `Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) ` `Delta_(r)^(@)G^(@)=-RTInK_(eq)` Consider the following reaction : `CO(g)+2H_(2)(g)toCH_(3)OH(g)` Given `Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol` `Delta_(r)H^(@)(CO,g)=-114KJ//mol`` s^(@)(CH_(3)OH,g)=240J//mol-k,` `S^(@)(H_(2)g)=198J//mol-K``C_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1)``C_(p.m(CO)=29.4J//mol-K` `C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K` `and In ((320)/(300))=0.06,"all data at"300K.` `Delta_(r)H^(@) `at 300K for the reaction is :A. `-87KJ/mol`B. `87KJ//mol`C. `172J//mol-K`D. none of these |
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Answer» Correct Answer - a |
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| 2906. |
For the reaction `,3O_(2)rarr 2O_(3),DeltaH=+ve`. We can say that `:`A. Ozone is more stable then oxygenB. Ozone is less stable then oxygen and ozone decomposes forming oxygen readilyC. Oxygen is less stable than ozone and oxygen decomposes forming ozone readilyD. None of the above |
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Answer» Correct Answer - B `3O_(2)rarr 2O_(3),Delta H=+ve` In endothermic reaction reactant is more stable than product. |
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| 2907. |
Which of the following reaction is endothermic?A. `CaCO_(3)(s)rarrCaO(s)+CO_(2)(g)`B. `Fe(s)+S(s)rarr FeS(s)`C. `NaOH(aq)+HCl(aq)rarr NaCl(aq)+H_(2)O(l)`D. `CH_(4)(g)+2O_(2)(g)+2H_(2)O(l)` |
| Answer» Correct Answer - A | |
| 2908. |
For the reaction `H_(2)+I_(2)hArr2NHI`, `DeltaH=12.40 kcal`. The heat of formation `(DeltaH)` of `HI` isA. 12.40 KcalB. `-12.40` KcalC. `-6.20` KcalD. 6.20 Kcal |
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Answer» Correct Answer - D `(1)/(2)H_(2(g))+(1)/(2)I_(2(g))rarr HI_((g)) , Delta H_(f) = ?` |
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| 2909. |
In the Born- Haber calculation of the lattice enthalpy of LiF from its elements , which process is exothermic?A. Dissociation energy of `F_(2)(g)`B. electron gain enthalpy of F(g)C. ionization ebergy of Li(g)D. sublimation erergy of Li(s) |
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Answer» Correct Answer - b |
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| 2910. |
Which one of the reaction is an exothermic reactionA. `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)`B. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`C. `2HgO(s)rarr 2Hg(s)+O_(2)(g)`D. None of these |
| Answer» Correct Answer - D | |
| 2911. |
The formation of water from `H_(2)(g)` and `O_(2)(g)` is an exothermic process because :A. The chemical energy of `H_(2)(g)` and `O_(2)(g)` is more than that of waterB. The chemical energy of `H_(2)(g)` and `O_(2)(g)` is less than that of waterC. The temperature of `H_(2)(g)` and `O_(2)(g)` is higher than that of waterD. The temperature of `H_(2)(g)` and `O_(2)(g)` is lower than that of water |
| Answer» Correct Answer - A | |
| 2912. |
For the process, `H_(2)O(l,T,K,P "bar")to H_(2)O(g,TK,P"bar")` Identify the option which is not correct.A. `DeltaG=0 if P, represent vapour pressure of `H_(2)O` at TK.B. `DeltaGlt0 "if P represent vapour pressure of `H_(2)O` at temperature greater than TK.C. `DeltaGgt0` if P is greater than vapour pressure of `H_(2)O` TK.D. Only A and C are correct. |
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Answer» Correct Answer - B |
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| 2913. |
Which of the following are applicable for a thermochemical equation ? It tells `:`A. It tells about physical state of reactants and productsB. It tells whether the reaction is spontaneousC. It tells whether the reaction is exothermic or endothermicD. It tells about the allotropic form (if any) of the reactants |
| Answer» Correct Answer - B | |
| 2914. |
Standard enthalpy of formation of `N_(2)O_(5)` is `-100 kcal//mol` and standard entorpy of `N_(2),O_(2) " and " N_(2)O_(5)` are 35, 40 and 115 kcal respectively, then `Delta_(r)G^(@)` of following reaction of `227^(@)` C will be : `2N_(2)+5O_(2)to2N_(2)O_(5)`A. `-80 kcal`B. `-180 kcal`C. `-1800 kcal`D. `+1800 kcal` |
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Answer» Correct Answer - B |
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| 2915. |
The correct thermochemical equation is :A. `C+O_(2)rarr CO_(2) , Delta H =-94` KcalB. `C+O_(2)rarr CO_(2) , Delta H = +94.0` KcalC. `C(s)+O_(2)(g)rarr CO_(2)(g) , Delta H = -94` KcalD. `C(s)+O_(2)(g)rarr CO_(2)(g) , Delta H = +94` Kcal |
| Answer» Correct Answer - C | |
| 2916. |
Calculate the enthalpy change where the standard heat of formation for gaseous `NH_(3)` is -11.02 kcal/mol at 298 K. The reaction given is `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) rarr NH_(3)(g)` |
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Answer» `Delta_(r)H^(Theta)=sum Delta_(f)H_("Product")^(Theta) - sum Delta_(f)H_("Reactant")^(Theta)` `=[Delta_(f)H^(Theta)(NH_(3))]-[(1)/(2)Delta H_(f)^(@)(N_(2))]+[(3)/(2)Delta H_(f)^(@)(H_(2))]` `=[-"11.02 kcal mol"^(-1)]-[(1)/(2)xx(0)]+[(3)/(2)xx(0)]` `= -"11.02 kcal mol"^(-1)-0` `= -"11.02 kcal mol"^(-1)` The standard enthalpies of formation of element in its reference state is taken as zero, therefore `Delta_(f)H^(Theta)` for `N_(2) and H_(2)` has been taken as zero. |
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| 2917. |
The enthalpy changes of formation of the gaseous oxides of nitrogen `(N_(2)O` and `NO)` are positive because of `:`A. The high bond energy of the nitrogen moleculeB. The high electron affinity of oxygen atomsC. The high electron affinity of nitrogen atomsD. The tendency of oxygen to form `O^(2-)` |
| Answer» Correct Answer - A | |
| 2918. |
Given the bond energies of `H - H` and `Cl - Cl` are `430 kJ mol^(-1)` and `240 kJ mol^(-1)`, respectively, and `Delta_(f) H^(@)` for `HCl` is `- 90 kJ mol^(-1)`. Bond enthalpy of `HCl` isA. `245 kJ mol^(-1)`B. `290 kJ mol^(-1)`C. `380 kJ mol^(-1)`D. `425 kJ mol^(-1)` |
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Answer» Correct Answer - D The enthalpy of formation of `HCl` is the enthalpy change when 1 mol of `HCl`is the enthalpy from its elements in their stantard states: `(1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) rarr HCl (g) , Delta_(g) H^(@) = - 90 kJ mol^(-1)` According to Eq. , we have `Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalpies"_("products")` `((1)/(2) Delta_(H - H) H^(@) + (1)/(2) Delta_(Cl - Cl) H^(@)) - Delta_(H - Cl) H^(@)` `- 90 = (1)/(2) (430) + (1)/(2) (240) - Delta_(H - Cl) H^(@)` `:. Delta_(H - Cl) H^(@) = (215) + (120) + (90) = 425 kJ mol^(-1)` |
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| 2919. |
Which respresent an exothermic reaction :-A. `2HgO(s)+180 K J rarr 2Hg(l)+O_(2)(s)`B. `N_(2)O(g)+C(s)rarr CO(g)+N_(2)(g)-131 K J`C. `N_(2)(g)+O_(2)(s)rarr 2NO(g) , Delta E = +181 K J`D. `C_(2)H_(2)(g)+2H_(2)(g)rarr C_(2)H_(6)(g) , Delta E = -314 K J` |
| Answer» Correct Answer - D | |
| 2920. |
molar standard enthalpy of combusion of ethanol is `-1320` kJ, malor standard enthalpy of formation of water and `CO_(2)(g)` are `-286` kJ//mole and `-393` kJ//mole respectively. Calculate the magnitude of molar standard enthalpy of formation of ethanol. |
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Answer» Correct Answer - 324 |
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| 2921. |
Standard heat of formation of `CH_(4)(g), CO_(2)(g)` and `H_(2)O(l)` are -75, -393.5, -286 kJ respectively. Find out the change in enthalpy for the reaction. `CH_(4)(g)+2O_(2)(g)rarr CO_(2)(g)+2H_(2)O(l)` |
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Answer» `CH_(4)(g)+2O_(2)(g)rarr CO_(2)(g)+2H_(2)O(l)` `Delta H_("of reaction")=[Sigma Delta H_(f)^(@)" of product" - Sigma Delta H_(f)^(@)" of reactant"]` `Delta H^(@)[(-393.5)+2(-286)]-[(-75)+2(0)]` `Delta H^(@)=(-965.5)-(-75)=-890.5 kJ` |
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| 2922. |
Heat capacity of water is `18 cal-"degree"^(-1)-mol^(-1)`. The quantity of heat needed to rise temperature of 18 g water by `0.2^(@)C` is x cal. Then amount of `CH_(4(g))` to be burnt to produce X cal heat is `(CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O, Delta H = -200 K.Cal)`A. `1.8 xx 10^(-3) mol`B. `3.6 xx 10^(-5) mol`C. `0.0288 g`D. `0.288 mg` |
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Answer» Correct Answer - D `Delta U = Zx theta(M)/(W), delta H = Delta U + Delta nRT, Q = mst` |
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| 2923. |
For an ideal monoatomic gas, an illustration of three different paths A,(B+C) and (D+E) from an initial state `P_(1),V_(1),T_(1)` to a final state `P_(2),V_(2),T_(1)` is shown in the given figure . Path A represents a reversible isothermal from `P_(1)V_(1)" to "P_(2),V_(2)`, path (B+C) represent a reversible adiabatic expansion (B) from `P_(1),V_(1),T_(1)" to " P_(3),V_(2),T_(2)` followed by reversible heating of the gas at constant volume (C) from `P_(3),V_(2),T_(2) " to " P_(2),V_(2),T_(1)`. Path (D+E) represents a reversible expansion at constant pressure `P_(1)(D)` from `P_(1),V_(1),T_(1) " to "P_(1),V_(2),T_(3)` followed by a reversible cooling at constant volume `V_(2)(E)` form `P_(1),V_(2),T_(3) " to "P_(2),V_(2),T_(1)` What is `DeltaS` for (D+E)?A. zeroB. `int_(T_(3))^(T_(1))(C_(V)(T))/(T)dt`C. `-nR" In"(V_(2))/(V_(1))`D. `nR" In"(V_(2))/(V_(1))` |
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Answer» Correct Answer - d |
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| 2924. |
Identify the correct statement regarding entropyA. At absolute zero temperature , entropy of a perfectly crystalline substance is taken to be zeroB. At absolute zero temperature, the entropy of a perfectly crystalline substance is positiveC. At absolute zero temperature , the entropy of all crystalline substance is to be zero.D. At `0^(@)C` , the entropy of a perfectly crystalline substance is taken to be zero. |
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Answer» Correct Answer - A "At absolute zero temperature entropy of a perfectly crystalline substance is taken to be zero." It is called third law of thermodynamics. |
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| 2925. |
For the reaction , `H_(2)(g)+1//2O_(2)(g)=H_(2)O(l), Delta C_(p)=7.63 "cal/deg" , Delta H_(25^(@)C)=68.3` Kcal, what will be the value (in Kcal) of `Delta H` at `100^(@)C` :A. `7.63xx(373-298)-68.3`B. `7.63xx10^(-3)(373-298)-68.3`C. `7.63xx10^(-3)(373-298)+68.3`D. `7.63xx(373-298)+68.3` |
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Answer» Correct Answer - C `(Delta H_(2)-Delta H_(1))/(T_(2)-T_(1))=Delta C_(p)` `Delta H_(2)=Delta C_(P)(T_(2)-T_(1))+Delta H_(1))` `Delta H_(2)=7.63 (100-25)+68.3` |
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| 2926. |
Which of the following value of `DeltaH_(f)^(@)` represent that the product is least stable ?A. `-94` K calB. `-231.6` K calC. `+21.4` K calD. `+64.8` K cal |
| Answer» Correct Answer - D | |
| 2927. |
Standard Gibbs energy of reaction `(Delta_(r)G^(@))` at a certain temperature can be completed as `Delta_(r)G^(@)=Delta_(r)H^(@)-Tdelta_(r)S^(@)` and the change in the value of `Delta_(r)H^(@)` and `Delta_(r)S^(@)` for a reaction with temperature can be computed as follows: `Delta _(r) H_(T_(2))^(@)-Delta_(r)C_(p)^(@)(T_(2)-T_(1))` `Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)In ((T_(2))/(T_(2)))` `Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) ` `Delta_(r)^(@)G^(@)=-RTInK_(eq)` Consider the following reaction : `CO(g)+2H_(2)(g)toCH_(3)OH(g)` Given `Delta_(r)H^(@)(CH_(3)Oh,g]=-201KJ//mol` `Delta_(r)H^(@)(CO,g)=-114KJ//mol`` s^(@)(CH_(3)OH,g)=240J//mol-k,` `S^(@)(H_(2)g)=198J//mol-K``C_(p.m)^(@)(h_(2))=28.8JK^(-1)mol^(-1)``C_(p.m(CO)=29.4J//mol-K` `C_(p.m)^(@)(CH_(3_)OH)=44J//mol-K` `and In ((320)/(300))=0.06,"all data at"300K.` `Delta_(r)H^(@) ` at 320K is :A. `-288.86KJ//mol`B. `-289.1KJ//mol`C. `-87.86KJ/mol`D. none of these |
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Answer» Correct Answer - c |
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| 2928. |
According to the following reaction `C(S)+1//2O_(2)(g)rarr CO(g), Delta H=-26.4` KcalA. CO is an endothermic compoundB. CO is an exothermic compoundC. The reaction is endothermicD. None of the above |
| Answer» Correct Answer - B | |
| 2929. |
Which of the following represents an exothermic reaction :-A. `N_(2)(g)+O_(2)(g)rarr 2NO(g), Delta H=180.5 KJ`B. `H_(2)O(g)+C(s)rarr CO(g)+H_(2)(g), Delta E=131.2KJ`C. `2HgO(s)+180.4 KJ rarr 2Hg(l)+O_(2)(g)`D. `2Zn(s)+O_(2)(g)rarr 2ZnO(s), Delta E = -693.8 KJ` |
| Answer» Correct Answer - D | |
| 2930. |
Calculate at the temperature `t = 16 ^@C` : (a) the root mean square velocity and the mean kinetic energy of an oxygen molecule in the process of translational motion , (b) the root mean square velocity of the molecules `eta = 1.50` times ? |
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Answer» (a) From the equipartition theorem `overline epsilon = (3)/(2) k T = 6 xx 10^-21 J` , and `v_(rms) = sqrt((3 k T)/(m)) = sqrt((3 RT)/(M)) = 0.47 km//s` (b) In equilibrium the mean kinetic energy of the droplet will be equal to that of a molecule. `(1)/(2) (pi)/(6) d^3 rho v_(rms)^2 = (3)/(2) kT` or `v_(rms) = 3 sqrt((2 kT)/(pi d^3 rho)) = 0.15 m//s`. |
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| 2931. |
From the reaction P(White) `rarr` P(Red) `Delta H =-18.4KJ`, it follows that :-A. Red P is readily formed from white PB. White P is readilly formed from red PC. White P can not be converted to red PD. White P can be converted into red P and red P is more stable |
| Answer» Correct Answer - D | |
| 2932. |
A 10.0 g piece of gallium (m=69.7) at `25.0^(@)` c is placed in 10.0 g of `H_(2)O"at" 55.0^(@)c` what is the final temperature when this system comes to equilbium ? (Assume the specific heat capacity of liquid Ga is the same as that of solid `Ga =0.37jg^((-1))"^(@)C^(-1)).`A. `35.0^(@)C`B. `38.1^(@)C`C. `41.8^(@)C`D. `52.6^(@)`C |
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Answer» Correct Answer - d |
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| 2933. |
A mass M = 15 g of nitrogen is enclosed in a vessel at temperature T = 300 K. What amount of heat has to be transferred to the gas to increase the root-mean-square velocity of molecules 2 times ? |
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Answer» Here `C_V = (5)/(2) (m)/(M) R(1 = 5 here)` `m` = mass of the gas, `M` = molecular weight, If `v_(rms)` increases `eta` times, the temperature will have increased `eta^2` times. This will require (neglecting expansion of the vessel) a heat flow of amount `(5)/(2)(m)/(M) R(eta^2 - 1) T = 10 k J`. |
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| 2934. |
Since the enthalpy of elements in their natural state is taken to be zero, the heat of formation `(Delta_(f)H)` of compoundsA. Is always negativeB. Is a always positiveC. is zeroD. May be positive or negative |
| Answer» Correct Answer - D | |
| 2935. |
For an ideal monoatomic gas, an illustration of three different paths A,(B+C) and (D+E) from an initial state `P_(1),V_(1),T_(1)` to a final state `P_(2),V_(2),T_(1)` is shown in the given figure . Path A represents a reversible isothermal from `P_(1)V_(1)" to "P_(2),V_(2)`, path (B+C) represent a reversible adiabatic expansion (B) from `P_(1),V_(1),T_(1)" to " P_(3),V_(2),T_(2)` followed by reversible heating of the gas at constant volume (C) from `P_(3),V_(2),T_(2) " to " P_(2),V_(2),T_(1)`. Path (D+E) represents a reversible expansion at constant pressure `P_(1)(D)` from `P_(1),V_(1),T_(1) " to "P_(1),V_(2),T_(3)` followed by a reversible cooling at constant volume `V_(2)(E)` form `P_(1),V_(2),T_(3) " to "P_(2),V_(2),T_(1)` What is `DeltaS` for path (A)?A. `nr" In"(V_(2))/(V_(1))`B. `-nr" In"(V_(2))/(V_(1))`C. zeroD. `nR(V_(2)-V_(1))` |
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Answer» Correct Answer - a |
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| 2936. |
The enthalpy of a reaction at 273 K. is -3.57 KJ. What will be the enthalpy of reaction at 373 K if `DeltaC_(p)` = zero :-A. `-3.57`B. ZeroC. `-3.57xx(373)/(273)`D. `-375` |
| Answer» Correct Answer - A | |
| 2937. |
Given the enthalpy changes: `A+BtoC DeltaH=-35KJxxmol^(-1)` `A+DtoE+FDeltaH=+20KJxxmol^(-1)``Fto C+EDeltaH=+15KJxxmol^(-1)` What is `DeltaH`for the reaction `2A+B+Dto2F` ?A. `0KJ.mol^(-1)`B. `-30KJ.mol^(-1)`C. `-40KJ.mol^(-1)`D. `-70KJxxmol^(-1)` |
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Answer» Correct Answer - b |
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| 2938. |
Given that `C+O_(2)rarrCO_(2),DeltaH^(@)=-xKJ` and `2CO+O_(2)rarr2CO_(2),DeltaH^(@)=-yKJ` The enthalpy of formation of carbon monoxide will beA. `y-2x`B. `2x-y`C. `(y-2x)/(2)`D. `(2x-y)/(2)` |
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Answer» Correct Answer - C `C + O_(2) to CO_(2),` `DeltaH^(@) =- x KJ` …(i) ON reversing given second equation we get, `2CO_(2) to 2CO + O_(2) , DeltaH^(@) = + y KJ` or `CO_(2) to CO + 1//2O_(2), DeltaH^(@) =+ y//2 KJ " "…(ii)` From Eqs. (i) and (ii) (by additon) `C+(1)/(2)O_(2) to CO,` `DeltaH^(@) =(y)/(2) -x =(y-2x)/(2) KJ` |
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| 2939. |
For the reactions, (i) `H_(2)(g)+Cl_(2)(g)rarr 2HCl(g)+ xKJ` (ii) `H_(2)(g)+Cl_(2)(g)rarr 2HCl(l)+ yKJ` Which one of the following statement is correct :A. x gt yB. x lt yC. x = yD. More data required |
| Answer» Correct Answer - B | |
| 2940. |
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using `dw=-P_("ext")dV` while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since `P=(nRT)/(V)`,so, `w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT " "In((V_(f))/(V_(i)))` Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. Two samples (initially under same states) of an idea gas are first allowed to expand to doubletheir volume using irreversible isothermal expansion against constant external pressure, then samples are turned back to their original volume first by reversible process having equation `PV^(2)`= constant then: A. final temperature of both samples will be equalB. final temperature of first sample will be greater than of second sampleC. Final temperature of second sample will be greater than of first sampleD. none of the above |
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Answer» Correct Answer - c |
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| 2941. |
Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid`/ /`base is completely neutralized by base `//`acid in dilute solution . For Strong acid and strong base neutralization net chemical change is`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)` `Delta_(r)H^(@)=-55.84KJ//mol``DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base `DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)` If enthalpy of neutralization of `CH_(3)COOH` by NaOH is -49.86KJ`//`mol then enthalpy of ionization of `CH_(3)COOH` is:A. 5.98KJ/molB. `-5.98KJ//mol`C. 105.7KJ/molD. none of these |
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Answer» Correct Answer - a |
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| 2942. |
Enthalpy of neutralization of NaOH with `H_(2)SO_(4)` is `-57.3KJeq^(-1)` and with ethanoic acid is `-55.2KJ.eq^(-1)` Which of thefollowing is the best explanation of this difference?A. Ethanioc acid is a weak acid and thus requires less NaOH for neutralization .B. Ethanoic acid is only partly ionised , neutralization is therefore incomplete.C. Ethanoic acid is monobasic while `H_(2)SO_(4)` Is dibasic.D. Some heat is used to ionize ethanioc acid completely. |
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Answer» Correct Answer - d |
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| 2943. |
Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid /base is completely neutralized by base `//`acid in dilute solution . For Strong acid and strong base neutralization net chemical change is`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)` `Delta_(r)H^(@)=-55.84KJ//mol``DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base `DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)` What is `DeltaH^(@)` for complate neutralization of strong diacidic base `A(OH)_(2)by HNO_(3)`?A. `-55.84KJ`B. `-111.68KJ`C. `55.84KJ`D. none of these |
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Answer» Correct Answer - b |
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| 2944. |
Calculate the enthalpy of isomerization of ethanol to dimethy ether: `C_(2)H_(5)OH(l)toCH_(3)OCH_(3)(g)` Given : Enthalpy of vaporisation of equal =41 KJ/mole bond enthalpies C-C=348KJ/mole C-H=415KJ/mole C-o=352KJ/mole O-h=463KJ/moleA. 65KJ/moleB. 25KJ/moleC. 125KJ/moleD. 85KJ/mole |
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Answer» Correct Answer - d |
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| 2945. |
Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid `//`base is completely neutralized by base `//`acid in dilute solution . For Strong acid and strong base neutralization net chemical change is`H^(+) (aq)+OH^(-)(aq)to H_(2)O(l)` `Delta_(r)H^(@)=-55.84KJ//mol``DeltaH_("ionization")^(@)` of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base `DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O)` under same conditions ,how many mL of 0.1 m NaOH and 0.05 M `H_(2)A` (strong diprotic acid ) solution should be mixed for a total volume of 100mL to producce the hight rise in temperature ?A. `25:75`B. `50:50`C. `75:25`D. `66.66:33.33` |
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Answer» Correct Answer - b |
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| 2946. |
Heat evolved in the reaction `H_(2)+Cl_(2) rarr 2HCl` is 182 kJ Bond energies H- H = 430 kJ/mole, `Cl-Cl = 242kJ//"mole"`. The H-Cl bond energy isA. `245 "KJ mol"^(-1)`B. `427 "KJ mol"^(-1)`C. `336 "KJ mol"^(-1)`D. `154 "KJ mol"^(-1)` |
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Answer» Correct Answer - B `H_(2)+Cl_(2)rarr 2HCl , Delta H = -182 kJ` `Delta H=Sigma(B.E.)_(R )-Sigma (B.E.)_(P)-182 =(430+242)-2xx(B.E.)_(H-Cl)` `therefore (B.E.)_(H-Cl)=427 KJ mol^(-1)` |
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| 2947. |
The coefficients in a balanced themochemical equation refer to the number of _______ of reactants and products involved in the reaction.A. molesB. moleculesC. volumesD. All of these |
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Answer» Correct Answer - A In the thermodynamics interpretation of an equation, we never interpret the coefficients as the number of molecules. Thus, it is acceptable to write coefficients as fractions rather than integers, wherever necessary. |
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| 2948. |
Select the incorrec option:A. Specific volume and molar heat capacity are intensive properties:B. Change in internal energy for an ideal gas for and isobaric process is expected as `nC_(v)(T_(2)-T_(1)).`C. Thermodyamics can predict rate at which process will take place.D. Free expansion is an irreversible process. |
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Answer» Correct Answer - C |
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| 2949. |
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compoundA. is always negativeB. is always positiveC. may be positive or negativeD. is never negative |
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Answer» Correct Answer - C The enthalpy of formation of a compound may be positive or negative as it can be exothermic or endothermic process. |
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| 2950. |
For an isothernal free expansion of an ideal gas correct question option is:A. `q=0,w=0, deltaH ne 0`B. `q=0,w=0, deltaH = 0`C. `q ne 0,w ne 0, deltaH ne 0`D. `q = 0,w = 0, deltaH = 0` |
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Answer» Correct Answer - D |
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