Chemical KineticsNEET Chemistry · Class 12 · NCERT Chapter 3

Solution

For an exothermic reaction, the equilibrium constant $K$ decreases with increasing temperature. Since $T_2>T_1$, $K_P>K_P^{\prime }$. This follows from the van't Hoff equation, which relates the change in equilibrium constant to the change in temperature for exothermic reactions, so option (a) is correct.

Solution

MgCO3(s)⟶MgO(s)+CO2(g) Moles of MgCO3=20 84=0.238 mol From abov e equation. 1 mole MgCO3 gives 1 mole MgO ∴ 0.238 mole MgCO3 will give 0.238 mole MgO =0.238×40 g=9.523 g MgO Practical yield of MgO=8 g MgO ∴% Purity= 8 9.523×100=84%

Solution

For zero-order kinetics (units of k indicate zero order): [B] = k·t = 0.6×10⁻³ × 20 × 60 = 0.72 M.

Solution

Rate Time 0.04 10 0.03 20 For first order reaction r ∝ conc. r1 r2 = C1 C2 = 4 3 ∴ k = 2.303 t2−t1 log C1 C2 ⇒ 0.693 t1 2⁄ = 2.303 20−10log 4 3 On solving t1 2⁄ = 24.1 s

Solution

Catalyst is going to affect the activation energy of a chemical reaction. Activation energy is the minimum energy required to from activated complex or Transition state.

Solution

A catalyst decreases activation energies of both the forward and backward reaction by same amount, therefore, it speeds up both forward and backward reaction by same rate. Equilibrium constant is therefore not affected by catalyst at a given temperature. www.vedantu.com 14

Solution

1/ 2 –2 0.693t s econd 10 = For the reduction of 20 g of reactant to 5 g, two t 1/2 is required. ∴ –2 0.693t2 s e c o n d 10 =× = 138.6 second

Solution

The solution of this question is given by assuming step (i) to be reversible which is not given in question Overall rate = Rate of slowest step (ii) = k[X] [Y 2] ...(1) k = rate constant of step (ii) Assuming step (i) to be reversible, its equilibrium constant, 112 22eq eq 2 2 [X]k[ X ] k [ X ][X ]= ⇒ = ...(2) Put (2) in (1) Rate = 11 22eq 2 2kk [X ] [Y ] Overall order = 13 122+=

Solution

The half-life of a first-order reaction is independent of the initial concentration $[\text{A}{]}_0$, while the half-life of a second-order reaction is dependent on $[\text{A}{]}_0$. This is a fundamental difference between the two types of reactions, as described in NCERT XII chapter Chemical Kinetics, so option (b) is correct.

Solution

For a zero-order reaction, the half-life $t_{1/2}=\frac{[A{]}_0}{2k}$. When the initial concentration $[A{]}_0$ is doubled, the half-life also doubles, so option (b) is correct.

Solution

– d[N2] dt = 1 2 d[NH3] dt

Solution

Orbital n l n+l 4d 4 2 6 5p 5 1 6 5f 5 3 8 6p 6 1 7

Solution

𝑋𝑒𝐹4:𝑠𝑝3𝑑2: Square planar 𝑋𝑒𝐹6:𝑠𝑝3𝑑3: distorted octahedral 𝑋𝑒𝑂𝐹4:𝑠𝑝3𝑑2: square pyramidal 𝑋𝑒𝑂3:𝑠𝑝3: pyramidal

Solution

An increase in the concentration of reactants leads to a higher collision frequency, as more particles are available to collide. This does not affect activation energy, heat of reaction, or threshold energy, so option (d) is correct.

Solution

The reaction $2\text{Cl}(g)\to {\text{Cl}}_2(g)$ involves the formation of a bond, which releases energy, making ${\Delta }_{r}H<0$. Additionally, the reaction decreases the number of gas molecules, reducing entropy, so ${\Delta }_{r}S<0$. Therefore, option (d) is correct.

Solution

For a first-order reaction, $t=\frac{2.303}{k}\log \left(\frac{[A{]}_0}{[A]}\right)$. Substituting, $t=\frac{2.303}{4.606\times 10^{-3}}\log \left(\frac{2.0}{0.2}\right)=\frac{2.303}{4.606\times 10^{-3}}\log 10=\frac{2.303}{4.606\times 10^{-3}}\times 1=500\text{s}$. Therefore, the correct option is (d).

Solution

The enthalpy of reaction is negative, indicating an exothermic reaction, and the enthalpy of activation is positive, indicating the energy barrier. The correct potential energy profile should show a lower energy for the products than the reactants, with an activation energy of 9.6 kJ mol${}^{-1}$. Option (a) correctly represents this profile.

Solution

The slope of the Arrhenius plot $\ln k$ vs. $\frac{1}{T}$ is given by $-\frac{E_a}{R}$. Given the slope is $-5\times 10^3\text{K}$ and $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$, we have $-\frac{E_a}{8.314}=-5\times 10^3\Rightarrow E_a=5\times 10^3\times 8.314=41570{\text{J mol}}^{-1}=41.57{\text{kJ mol}}^{-1}\approx 41.5{\text{kJ mol}}^{-1}$. However, the closest option is (b) 83.0 kJ mol${}^{-1}$, which suggests a possible typo in the options. The correct value should be 41.5 kJ mol${}^{-1}$.

Solution

• For zero order reaction r = k[A]0 r = k (constant) hence, ‘y’ as ‘rate’ and ‘x’ as concentration will give desired graph. • For first order reaction ½ 0.693t (constant)k= hence, ‘y’ as ‘t½’ and ‘x’ as concentration will give desired graph.

Solution

List – I List – II (Products formed) (Reaction of carbonyl compound with) (a) Cyanohydrin → HCN (b) Acetal → Alcohol (c) Schiff's base → RNH2 (d) Oxime → NH2OH (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)

Solution

For first order reaction, 0A]2.303K log t [A] [= ; where A0 is the initial concentration of reactant A. A0 = 0.1 M A = 0.001 M t = 5 minute K = 22.303 0.1 2.303log log105 0.001 5 = = 2.303 25 × K = 0.9212 min–1 - 36 - NEET (UG)-2022 (Code-Q1)

Solution

The magnetic force on a current-carrying wire is given by $\vec{F}=I\vec{L}\times \vec{B}$. Here, $\vec{L}=L\hat{i}$ and $\vec{B}=2\hat{i}+3\hat{j}-4\hat{k}$. The cross product $\vec{L}\times \vec{B}=L\hat{i}\times (2\hat{i}+3\hat{j}-4\hat{k})=L(0+3\hat{k}+4\hat{j})=L(4\hat{j}+3\hat{k})$. The magnitude of the force is $F=IL\sqrt{4^2+3^2}=5IL$, so option (d) is correct.

Solution

Assertion A is false because a reaction cannot have zero activation energy; some energy is always required to break bonds and form products. Reason R correctly defines activation energy, so option (a) is correct.

Solution

Heterogeneous catalysis involves a catalyst in a different phase from the reactants. In the Haber process, finely divided iron (solid) catalyzes the reaction between gaseous dinitrogen and dihydrogen, making option (a) the correct example of heterogeneous catalysis.

Solution

The rate law is given by $r=k[A{]}^2[B]$. When $[A]$ is tripled, the new rate $r^{\prime }=k(3[A]{)}^2[B]=9k[A{]}^2[B]=9r$. Therefore, the initial rate increases by a factor of nine, so option (d) is correct.

Solution

The Arrhenius equation is given as aE RTk Ae − = ∴ aEln k ln A RT=− ln k v/s 1 T gives a straight line graph with slope = aE R− and intercept = ln A

Solution

To calculate value of Ea Equation used is a2 1 1 2 Ek 11log –k 2.303R T T ⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Hence Ea can be calculated if value of rate constant k is known at two different temperatures T 1 and T2.

Solution

a2 1 1 2 Ek 11log = –k 2.303R T T ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ aE4 1 1log = –1 2.303R 300 330 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ Ea = ( )( )log 4 2.303 8.314 300 330 30 × × × × = 3.804 × 104 J/mol = 38.04 kJ/mol

Solution

For 1ˢᵗ order reaction

$kt=2.303\log \frac{A_0}{A_t}$ $A_0$ = initial concentration

$A_t$ = Final concentration

$t_{99.9\%}=10t_{1/2}$

$t_{99.9\%}=10\times 1\text{ minute}=10\text{ minutes}$

Solution

Sol. $K_C=\frac{k_f}{k_b}=\frac{1}{2500}$

$K_P=K_C(RT{)}^{\Delta n_g}(\Delta n_g=2-1=1)$

$=\frac{1}{2500}\times 0.0831\times 1000$

$=0.033$

Solution

Sol. $k=0.03{\text{s}}^{-1}$
$t=\frac{2.303}{k}\log \frac{a}{a-x}$
$=\frac{2.303}{0.03}\log \frac{7.2}{0.9}$
$=\frac{2.303}{0.03}\log 8$
$=\frac{2.303}{0.03}\times 3\times \log 2$
$=\frac{2.303}{0.03}\times 3\times 0.301$
$=69.3\text{s}$