Question No 1
The question tests on simple application of Le Chateliers law=> If a Change is made to a system which is at equilibrium then th e equilibrium will shift in a direction to oppose or minimize the change
. 
The question emphasizes
to select a change that does not increase the rate of the forward reaction,
this statement is equivalent to saying
which change increases the rate of the backward direction.
The forward reaction is a exothermic process so to increase
the rate of backward reaction which is endothermic , we have to increase the
temperature. So the absolute option is C.
Lets look into option A.
Addition of a catalyst or increasing
the surface area of the of the catalyst by decreasing the size of the catalyst
pieces does not selectively increase the rate of backward or forward reaction
because the rate of both reactions are equally increased. So in this case also
the rate of forward reaction is increased , as well as the rate of the backward
reaction.
Consider option B.Increasing the pressure will try to shift
the equilibrium in a direction where the total number of molecules is lesser ,
so that the pressure can be reduced. In the Haber Process equation the number
of molecules on the reactant side is 4 ( one N2 and three H2) and the number of
molecules on the product side is 2.So
when the pressure is increased the forward process is favoured .
According to option D, removal of ammonia from the system as
it is formed will make the system to produce more ammonia, so the rate of the
forward direction has to be increased to
make more ammonia.
Question No 2 Reference to
a species which has a incomplete outer shell is shown here. This means a atom
or ion which has not eight electrons in its outer most shell .Let consider the options individually,BThe forward reaction is a exothermic process so to increase
the rate of backward reaction which is endothermic , we have to increase the
temperature. So the absolute option is C.
Lets look into option A.
Addition of a catalyst or increasing
the surface area of the of the catalyst by decreasing the size of the catalyst
pieces does not selectively increase the rate of backward or forward reaction
because the rate of both reactions are equally increased. So in this case also
the rate of forward reaction is increased , as well as the rate of the backward
reaction.Consider option B.Increasing the pressure will try to shift
the equilibrium in a direction where the total number of molecules is lesser ,
so that the pressure can be reduced. In the Haber Process equation the number
of molecules on the reactant side is 4 ( one N2 and three H2) and the number of
molecules on the product side is 2.So
when the pressure is increased the forward process is favoured . According to option D, removal of ammonia from the system as
it is formed will make the system to produce more ammonia, so the rate of the
forward direction has to be increased to
make more ammonia. Question No 2 Reference to
a species which has a incomplete outer shell is shown here. This means a atom
or ion which has not eight electrons in its outer most shell .Let consider the options individually,BThe forward reaction is a exothermic process so to increase
the rate of backward reaction which is endothermic , we have to increase the
temperature. So the absolute option is C.
Lets look into option A.
Addition of a catalyst or increasing
the surface area of the of the catalyst by decreasing the size of the catalyst
pieces does not selectively increase the rate of backward or forward reaction
because the rate of both reactions are equally increased. So in this case also
the rate of forward reaction is increased , as well as the rate of the backward
reaction.Consider option B.Increasing the pressure will try to shift
the equilibrium in a direction where the total number of molecules is lesser ,
so that the pressure can be reduced. In the Haber Process equation the number
of molecules on the reactant side is 4 ( one N2 and three H2) and the number of
molecules on the product side is 2.So
when the pressure is increased the forward process is favoured . According to option D, removal of ammonia from the system as
it is formed will make the system to produce more ammonia, so the rate of the
forward direction has to be increased to
make more ammonia. Question No 2 Reference to
a species which has a incomplete outer shell is shown here. This means a atom
or ion which has not eight electrons in its outer most shell .Let consider the options individually,BThe forward reaction is a exothermic process so to increase
the rate of backward reaction which is endothermic , we have to increase the
temperature. So the absolute option is C.
Lets look into option A.
Addition of a catalyst or increasing
the surface area of the of the catalyst by decreasing the size of the catalyst
pieces does not selectively increase the rate of backward or forward reaction
because the rate of both reactions are equally increased. So in this case also
the rate of forward reaction is increased , as well as the rate of the backward
reaction.Consider option B.Increasing the pressure will try to shift
the equilibrium in a direction where the total number of molecules is lesser ,
so that the pressure can be reduced. In the Haber Process equation the number
of molecules on the reactant side is 4 ( one N2 and three H2) and the number of
molecules on the product side is 2.So
when the pressure is increased the forward process is favoured . According to option D, removal of ammonia from the system as
it is formed will make the system to produce more ammonia, so the rate of the
forward direction has to be increased to
make more ammonia. Question No 2 Reference to
a species which has a incomplete outer shell is shown here. This means a atom
or ion which has not eight electrons in its outer most shell .Let consider the options individually,BF3 :
Question no 3
Ionization energy is the amount of energy needed to to
remove one mole of electron from the outermost shell of a species in the
gaseous state.When a electron is removed from a neutral species to form a unipositive ion
it is called as the first ionization energy. Removal of a electron from a
unipositive ion to form a dipositive ion is called as the second ionixation
energy,Erg: CaàCa+………………..1
This is not the equation for the second ionization energy but represents the sum of the first and second ionization enthalpy.And also always ionization energy is a endothermic process , so the sign of the enthalpy change should be positive. =>∆H o = +So the correct option is B.Question 4The question is easy to handle.
This reaction can be considered as both the equation for the formation of H2O ( water) or that of the combustion of H2(g) .You might not know whether the formation enthalpy of h2o is exothermic but you certainly know that the combustion enthalpy of h2 is exothermic.As both reaction are the same, the same amount of heat is liberated by hess’s law.So the correct option is D.Question 5Always when faced with such a question immediately try to analyse the data and draw a triangle of balanced equations inorder to apply Hess’s law and find the unknown value . CH3COCH3 +3O2 ----------à2CO2 + 3H2O
H2 + 1/2 O2 ----à H2O
C+ O2 --à CO2Now lets draw the triangle: –1786 kJ mol–1.
–286 kJ mol–1 X 3Unknown Y –394 kJ mol–1 X 2 3H2 + 2C + 7/2 O2
By doing the math:Y= –254 kJ mol–1So the correct answer is option CQuestion No 6Consider the shapes of each of the species given in the question. Question No 7The question is so simple only easy substitution and the rest is mathamatics .But remember all the quantities in the ideal gas equation have to be put in the SI unit before you substitute,So lets first do thatP(Pressure) = 102 kPa = 102000 PaT(absolute temperature)= 30 C = 30+273 K = 303 KN(moles) = mass/molar mass` = 0.56g/30gmol-1 =0.01866 mol molar mass of ethane (C2H4) = 30 gmol-1 PV=nRTV= Nrt/PV= M3=cm3So the correct option is
Question 8Just to remind , the properties of a ideal gas:The molecules have neither attraction or repulsion forces between them.The volume of a single molecule can be neglected compared to the total volume of the vessel.The collisions between the molecules are perfectly elastic.But we all know that ideal gas is a imaginary concept and such a gas does not really exist, but some real gases like the noble gases behave more ideally than other gases. So now our task is to select appropriate condition in which a real gas would more likely show ideal nature.Consider the temperature, the temperature has to be very high for a real gas to exhibit ideal nature, because then only 1.the molecules of gas will be moving very fast and chance for attraction between molecules is less2.According to Charles law when the temperature increases the volume also increases ,to the volume of the molecules is more negligle compared to that of the vessel.From the above explanation , you would certainly understand that the pressure of the system need to be low for the gas to show ideal nature, because at low pressure the total volume of the gaseous system is higher so the volume of the gas molecules is more negligible .Question 9A thorough understanding of the process by which Al is extracted from bauxite would be very much usefulful to tackle this question.Lets consider each options individually,
Option A:In this process Al2O3 is molten with cryolite , remember Al2O3 is a ionic compound.But AlCl3 is a covalent compound,Extraction of substances throught the method of electrolysis is not possible for covalent substances, because here ions can not be formed .Al2O3 dissociates into Al3+ and O2-.Al3+ is a cation and moves towards the negatively charged cathode,
Always remember that oxidation takes place at the anode and reduction takes place at the cathode.
Reaction at cathode:Al3+ + 3e -àAl
Reaction at anode :2O2- àO2 +2e
So Al3+ move towards the cathode and are reduced.
