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.**A collision between reactant molecules must occurs with a certain minimum energy before it is effective in yielding product molecules. This minimum energy is called activation energy E_a. Larger the value of activation energy, smaller the value of rate constant k. Larger is the value of activation energy, greater is the effect of temperature rise on rate constant k E_f= Activation energy of forward reaction E_b= Activation energy of backward reaction ∆H=E_f-E_b E_f= Threshold energy 344. If a reaction A+B→C is exothermic to the extent of 30 kJ mol^(-1) and the forward reaction has an activation energy of 249 kJ mol^(-1) the activation energy for reverse reaction in kJ mol^(-1) is

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.**The order of reaction is an experimentally determined quantity. It may be zero, positive, negative, or fractional. The kinetic equation of nth order reaction is k×t=1/((n-1)) [-1/(a-x)^(n-1) -1/a^(n-1) ] …(i) Half life of nth order reaction depends on the initial concentration according to the following relation: t_(1/2)∝1/a^(n-1) …(ii) The unit of the rate constant varies with the order but general relation for the unit of nth order reaction is Unit of k=[1/Conc]^(n-1)×Time^(-1) …(iii) The different rate law for nth order reaction may be given as: dx/dt=k[A]^n …(iv) Where A denotes the reactant 345. The unit of rate and rate constant are same for :

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.**Consider the following elementary reaction, 2A+B+C→ Products. All reactants are present in the gaseous state and reactant C is taken in excess 346. What is the rate expression of the reaction?

Solution

(c) Explanation is correct reason for statement.

(c) Explanation is correct reason for statement.

**Q4.**For the reaction: X(g)→Y(g)+Z(g), the following data were obtained at 30℃: Experiment [X] (mol L^(-1)) Rate (mol L^(-1) hr^(-1)) I 0.17 0.05 II 0.34 0.10 III 0.68 0.20 347. The rate constant of the above 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.**The reaction 2AX(g)+2B_2 (g)→A_2 (g)+2B_2 X(g) has been studied kinetically and on the basis of the rate law following mechanism has been proposed 2AX⇌A_2 X_2 (fast and reverse) A_2 X_2+B_2→A_2 X+B_2 X (slow) A_2 X+B_2→A_2+B_2 X (fast) Where all the reaction intermediates are gases under ordinary condition From the above mechanism in which the steps (elementary) differ considerably in their rates, the rate law is derived using the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the product of the molar concentration of each reaching species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In order to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step 348. Let the equilibrium constant of Step I be 2×〖10〗^(-3) mol^(-1) L and the rate constants for the formation of A_2 X and A_2 in Steps II and III are 3.0×〖10〗^(-2) mol^(-1) L〖 min〗^(-1) and 1×〖10〗^3 mol^(-1) L min^(-1) (all data at 25℃), then what is the overall rate constant (mol^(-2) L^2 min^(-1)) of the consumption of B_2?

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

**Q6**The rate law expression is given for a typical reaction. n_1 A+n_2 B→P as r=k[A]^n [B]^(n_2 ). The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by considering the slowest step, i.e., for S_N1 reaction r=k[RX] If the reaction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is considered, i.e., the rate of formation of intermediate is always equal to the rate of decomposition of the intermediate Consider the reaction: 349. If we increase the concentration of I_2 two times, then the rate of formation of HI will

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

**A secondary alkyl halide (A) hydrolyzes with alkali (B) in aqueous medium simultaneously via S_N 1 and S_N2 pathways with rate constants k_1 and k_2, respectively. From kinetic data, it was found that a plot of =(-1)/([A]) (d[A])/dt vs[B] is straight line with a slope equal to 2.7×〖10〗^(-4) L mol^(-1) min^(-1) and intercept equal to 1.02×〖10〗^(-3). Minimum initial concentration of [A]=0.2 M and [B], i.e., [ ]=0.5 M 350. The value of overall rate constant of the hydrolysis of A (in L mol^(-1) min^(-1)) is**

**Q7.**
Solution

**Q8.**If a unimolecular reaction, A(g)→ Products, takes place according to the mechanism Where k_1,k_(-1) and k_2 are the rate constants and P, A and A^* stand for product molecule, normal molecules of reactants and activated molecules of reactants respectively 351. Which of the following expressions are correct?

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.**The thermal decomposition of N_2 O_5 occurs as: 2N_2 O_5⟶4NO_2+O_2 Experimental studies suggest that rate of decomposition of N_2 O_5, rate of formation of NO_2 or rate of formation of O_2 all becomes double if concentration of N_2 O_5 is doubled. 352. The correct mechanism for the decomposition of N_2 O_5 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.