IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1.Half-life t_(1/2) of reaction is the time required for the concentration of reactant to derease by half, ie,
[A]_t=1/2 [A]
t_(1/2)=0.693/K
That is t_(1/2) is independent of initial concentration. Figure shows that typical variation of concentration of reactant exhibiting first order kinetics. It may be noted that though the major portion of the first order kinetics may be over in a finite time, but the reaction will never cease as the concentration of reactant will be zero only at infinite time
75% of a reaction of the first order was completed in 32 min. When was it half completed?
Solution
(d) Both statement and explanation are correct but the reason is different
(d) Both statement and explanation are correct but the reason is different
Q2.Note all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions. Let a reaction be represented in general terms by the scheme
Where k_1 and k_2 represent the rate constant for the forwards and reverse reactions respectively. The equilibrium constant for this reaction may be written as
k=[B]_∞/[A]_∞=k_1/k_2
The initial concentration of species A is [A]_0 and that of B is [B]_0. After a time t, let the concentration of species A be [A]_t and that of B be [B]_t. The total rate of change of [A]_t is given by
d[A]_(t/dt)=-k_1 [A]_t+k_2 [B]_t
d[A]_(t/dt)=-k_1 [A]_t+k_2 ([A]_0-[A]_t )
d[A]_(t/dt)=-(k_1+k_2 )([A]_t-k_2/(k_1+k_2 ) [A]_0 )
334. What is the k_c of the reaction 2C⇌2A?
:
Solution
(a) Nacl has fcc structure in which each Na^+ is surrounded by six ions and vic versa.In this octahedral arrangement, coordination, number of bothNa^+ and Cl^- is six for which radius ratio lies between 0.414 and 0.732. The radius ratio does not allow Cl^- ions to touch each other.
(a) Nacl has fcc structure in which each Na^+ is surrounded by six ions and vic versa.In this octahedral arrangement, coordination, number of bothNa^+ and Cl^- is six for which radius ratio lies between 0.414 and 0.732. The radius ratio does not allow Cl^- ions to touch each other.
Q3. The rate of reaction increases significantly with increase in temperature. Generally, rates of reactions are doubled for every 10° rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10° change in temperature
Temperature coefficient (μ)=(Rate constant of (T+10)℃)/(Rate constant at T℃)
Arrhenius gave an equation which describes arte constant k as a function of temperature
k=Ae^(-E_a/RT)
Where k is the rate constnat, A is the frequency factor or pre-exponetial factor, E_a is the activation energy, T is the temperature in Kelvin, and R is the universal gas constant.
Equation when expressed in logarithmic form becomes
logk=logA-E_a/(2.303 RT)
335. For a reaction E_a=0 and k=3.2×〖10〗^8 s^(-1) at 325 K. The value of k at 335 K would be
Solution
(c) Explanation is correct reason for statement.
(c) Explanation is correct reason for statement.
Q4. The rate constant for the decomposition of a certain reaction is described by the equation: logk (s^(-1) )=14-(1.25×〖10〗^4 K)/T 336. Pre-exponential factor for this reaction is
Solution
(a) Packing fraction in fcc = 74% Packing fraction in bcc = 67.9%
(a) Packing fraction in fcc = 74% Packing fraction in bcc = 67.9%
Q5.style="text-indent: -48px; white-space: pre;"The following data were observed for the following reaction at 25℃,CH_3 OH+(C_6 H_5 )_3 CCl→(C_6 H_5 )_3 C∙OCH_3+HCl
Set Initial concentration
(M)
[A]_0 [B]_0 Time (∆t) (min) Final concentration (M)
I
II
III 0.10
0.10
0.20 0.05
0.10
0.10 25
15
7.5 0.0033
0.0039
0.0077
337. Rates d[C]/dt in sets I, II and III are, respectively, (in M min^(-1)):
I II III
Solution
(d) Both statement and explanation are correct but reason for the statement is different
(d) Both statement and explanation are correct but reason for the statement is different
Q6The energy profile diagram for the reaction:
CO(g)+NO_2 (g)⇌CO_2 (g)+NO(g)
In given below:
338. The activation energy of the forward reaction is
Solution
(a) For fcc, Number of O^(2-) ions =8(corners)×1/8per corner share +6 (faces)×1/2per face share=4 Number of Si^(4+) ions =1/4×OV=1/4×4=1/unit cell Number Fe^(2+) ions =1/4×TV=1/4×8=2/unit cell ⇒Fe_2 SiO_4
(a) For fcc, Number of O^(2-) ions =8(corners)×1/8per corner share +6 (faces)×1/2per face share=4 Number of Si^(4+) ions =1/4×OV=1/4×4=1/unit cell Number Fe^(2+) ions =1/4×TV=1/4×8=2/unit cell ⇒Fe_2 SiO_4
Q7. The reaction S_2 O_8^( 2-)+3I^⊝→2SO_4^( 2-)+I_( 3)^⊖ is of first order both with respect to persulphate and iodide ions. Taking the initial concentration as a and b, respectively, and taking x as the concentration of the triodide at time t, a different rate equation can be written
Two suggested mechanisms for the reaction are:
339. The general different equation for the above reaction is
Solution
Q8.Consider the reaction represented by the equation:
CH_3 Cl (g)+H_2 O(g)→CH_3 OH(g)+HCl(g)
Those kinetic data were obtained for the given reaction concentration:
Initial conc (M) Initial rate of disappearance
[CH_3 Cl] [H_2 O] of CH_3 Cl(M s^(-1))
0.2 0.2 1
0.4 0.2 2
0.4 0.4 8
340. The rate law for the reaction will be
Solution
(c) The distance 0.204 nm comes out to be " a⁄√3 " which corresponds to the perpendicular distance between a corner of unit cell and the plane of the three adjacent corners (Fig. III)
(c) The distance 0.204 nm comes out to be " a⁄√3 " which corresponds to the perpendicular distance between a corner of unit cell and the plane of the three adjacent corners (Fig. III)
Q9.For the reaction: aA+bB→cC+dD
Rate =dx/dt=(-1)/a (d[A])/dt=(-1)/b (d[B])/dt=1/c (d[C])/dt=1/d (d[D])/dt
341. For reaction 3BrO^⊝→BrO_3^( ⊖)+2Br^⊖, the value of rate constant at 80℃ in the rate law for –d[BrO^⊝ ]/dt was found to be 0.054 L mol^(-1) s^(-1). The rate constant (k) for the reaction in terms of (d[BrO_3^( ⊝)])/dt is
Solution
(a) AB_2 has bcc structure, A^(2+) possess face centred cubic lattice. B^- ions occupy all the (100%) tetrahedral voids. Thus each A^(2+) is in contact with B^- and each B^- with 4A^(2+) ions.
(a) AB_2 has bcc structure, A^(2+) possess face centred cubic lattice. B^- ions occupy all the (100%) tetrahedral voids. Thus each A^(2+) is in contact with B^- and each B^- with 4A^(2+) ions.
Q10.The energy change accompanying the equilibrium reaction A⇌B is -33.0 kJ mol^(-1)
Assuming that pre-exponential factor is same for forward and backward reaction answer the following
342. The equilibrium constant k for the reaction at 300 K
Solution
(b) All factual statements
(b) All factual statements