Class 12th

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

Reaction occurring at the smallest time interval is known as instantaneous rate of reaction. For e.g the instantaneous rate of reaction at 40s is the rate of reaction during a small interval of time close to 40s. Volume changes during a small-time interval close to the 40s.

Instantaneous rate can be determined graphically by drawing a tangent on the curve

Instantaneous reaction= $\frac{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}}{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}$

Option B is incorrect since the line travels through the graph but does not link to any of the Y axis lines. As a result, option B is incorrect.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

Option (c) is the only one of the four assertions that is incorrect. In rate law expression, the order of reaction is equal to the sum of the power of concentration of the reactants.

xA + yB→zC

r = k (A)^x (B)^y 

Order x + y

The order of the reactions can also be a fraction. In a balanced chemical equation, the order of reaction may or may not be equal to the total of the stoichiometric coefficients of the reactants.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

In the given graph V is the volume in the Y axis and time is in X axis

Zn + dil.HCl→ZnCl₂ + H₂↑

Average rate = 40s

Average rate of reaction = $\frac{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{H}}_2}{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}$

$\frac{\mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}-\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}-\mathrm{}\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}}$

Analyzing the graph line where time and volume intersect. 

$\frac{{\mathrm{V}}_3-0}{40-0}$

$\frac{{\mathrm{V}}_3}{40}$

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option D

k = Ae -    $\frac{\mathbf{}\mathbf{E}\mathbf{a}}{\mathbf{R}\mathbf{T}}$             

From this equation

Kα $\frac{1}{\mathrm{E}\mathrm{a}}$

When Activation energy Ea decreases, rate constant k

 increases.

Hence, Rate constant increases exponentially with decreasing activation energy and increasing temperature. 

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option A

Arrhenius equation Ae -   $\frac{\mathbf{}\mathbf{E}\mathbf{a}}{\mathbf{R}\mathbf{T}}$

K = rate constant

A= frequency factor

Ea= Activation Energy

R= gas constant

T= temperature

ln k = lnA - $\frac{\mathbf{E}\mathbf{a}}{\mathbf{R}\mathbf{T}}$

When the temperature rises, ln falls.

∴ ln kv/s $\frac{1}{\mathbf{T}}\mathrm{}$ , is a negative slope.

ln A is intercepted by k, and its magnitude decreases with time.

∴ Negative slope is obtained.

When compared to the other possibilities, T is increasing over time, which is incorrect.

ln k =  $\frac{-\mathbf{E}\mathbf{a}}{\mathbf{R}}$ $\frac{1}{\mathbf{T}}\mathbf{}$  + lnA

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

Given:  

A (g) → B (g) + C (g) 

p_i= Initial pressure

(Time) t = 0

A (g) → B (g) + C (g) 

p_i        →0atm + 0atm

t, (p_i - x)atm

p_t = (p_i - x)atm + x + x = p_i + x

p_A = (p_i - x)

The value of x changes when it is substituted.

p_A = p_i - (p_t - p_i) = 2p_i - p_t

K =  $\frac{2.303}{\mathrm{t}}$ log $\frac{{\mathrm{p}}_{\mathrm{i}}}{2{\mathrm{p}}_{\mathrm{i}}-{\mathrm{p}}_{\mathrm{t}}}$

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option A

Activation Energy → The amount of energy required to overcome the obstacle and generate a product

The activation energy of a forward reaction can be seen in the [E_af = E₁ + E₂] (This is an endothermic reaction.

Therefore, [E_af > E_ab]   

The energy of the product is high, while the energy of the reactant is low.

The lower the energy, the more stable and positive the situation becomes.

Δ? = posistive

13.27 In the fission of ${}_{92}{}^{238}U$ , ${\beta }^{-}$ 10  particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}{}^{238}U+{}_0{}^{1}n\to {}_{58}{}^{140}Ce+{}_{44}{}^{99}Ru+10{}_{-1}{}^{0}e$

It is given that:

Mass of a ${}_{92}{}^{238}\mathrm{U}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_{1\mathrm{}}$ = 238.05079 u

Mass of a ${}_{58}{}^{140}\mathrm{C}\mathrm{e}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_2$ =139.90543 u

Mass of a ${}_{44}{}^{99}\mathrm{R}\mathrm{u}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_3$ = 98.90594 u

Mass of a neutron ${}_0{}^{1}n$ , $m_4$ = 1.00865 u

Q value of the above equation,

Q = $\left[m^{\text{'}}\left({}_{92}{}^{238}\mathrm{U}\right)+\mathrm{}\mathrm{m}\mathrm{\text{'}}({}_0{}^1\mathrm{n})-{\mathrm{m}}^{\mathrm{\text{'}}}({}_{58}{}^{140}Ce)-m^{\text{'}}({}_{44}{}^{99}Ru)-10m_e\right]c^2$

Where

m' = represents the corresponding atomic masses of the nuclei.

$m^{\text{'}}\left({}_{92}{}^{238}\mathrm{U}\right)$ = ${\mathrm{m}}_{1\mathrm{}}-\mathrm{}92m_e$

m'( ${}_{58}{}^{140}Ce)$ = ${\mathrm{m}}_2-\mathrm{}58m_e$

m'( ${}_{44}{}^{99}Ru)$  = $m_3-44m_e$

m'( ${}_0{}^{1}n)=m_4$

Substituting these values, we get

Q = $\left[{\mathrm{m}}_{1\mathrm{}}-\mathrm{}92m_e+m_4-{\mathrm{m}}_2+\mathrm{}58m_e-m_3+44m_e-10m_e\right]c^2$

= $\left[{\mathrm{m}}_{1\mathrm{}}+\mathrm{}{m}_4-m_3-{\mathrm{m}}_2\right]c^2$

= $\left[238.05079+1.00865-98.90594-139.90543\right]c^2$ u

=0.24807 $c^2$ u

= 231.077 MeV