CBSE 12 Class Exams Questions and Answers

# NCERT

Out of $NaCl$ and $AgCl$ , which one shows Frenkel defect and why?

S Sumit

shows Frenkel defect because of the small size of ion, it can easily fit in the interstitial sites while can not as it is bigger in size.

# NCERT

What is binary number

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N neha

A binary number uses only two digits zero and one. In a binary system, numbers are represented as follows

NUMBER AS INTEGER	NUMBER IN BINARY
0	00
1	01
2	10
3	11
4	100
5	101
6	110
7	111

# NCERT

Why is glass considered a super cooled liquid?

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R Ramesh

Ever observed an old glass window? It will be thicker at the base than the top. This shows that glass flows(at a very slow rate) and it can flow because it is amorphous.

# NCERT

3.20 b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (iii) 6 Ω

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R Rajeev Yadav

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Question 6. Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

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P pooja

test> <sc

# NCERT

16. Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit. Discuss it with your teacher.

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B Bipasha Kumar

It's called dependence and addiction. If the use or intake of alcohol is suddenly discontinued it leads to withdrawal symptoms.

# NCERT

Q 15)
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

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G Gautam harsolia

Given function is Given function is satisfies for the all real values of x case (i) k < 0 Hence, function is continuous for all values of x < 0 case (ii) x = 0 L.H.L at x= 0 R.H.L. at x = 0 L.H.L. = R.H.L. = f(0) Hence, function is continuous at x = 0 case (iii) k > 0 Hence , function is continuous for all values of x > 0 case (iv) k < 1 Hence , function is...

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4.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

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M manish

From the Arrhenius equation, ...................................(i) it is given that T1= 293 K T2 = 313 K Putting all these values in equation (i) we get, Activation Energy = 52.86 KJ/mol This is the required activation energy

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4.29 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is $4 \times 10 ^{10} s ^{-1}$ . Calculate k at 318K and Ea.

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M manish

We know that, for a first order reaction- Case 1 At temp. = 298 K = 0.1054/k Case 2 At temp = 308 K = 2.2877/k' As per the question K'/K = 2.7296 From Arrhenius equation, = 76640.096 J /mol =76.64 KJ/mol k at 318 K we have , T =318K A= Now After putting the calue of given variable, we get on takingantilog we get, k =...

# NCERT

4.28 The decomposition of A into product has value of k as $4.5 \times 10 ^3 s ^{-1}$ at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be $1.5 \times 10 ^4 s ^{-1}$ ?

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S seema garhwal

The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1. K1 = K2 = = 60 kJ mol–1 K2 =

# NCERT

4.27 The rate constant for the first order decomposition of $H_2O_2$ is given by the following equation:
$\log k = 14.34 - 1.25 \times 10 ^ 4K/T$ .

Calculate $E_a$ for this reaction and at what temperature will its half-period be 256 minutes?

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M manish

The Arrhenius equation is given by taking log on both sides, ....................(i) given equation, .....................(ii) On comparing both equation we get, activation energy half life () = 256 min k = 0.693/256 With the help of equation (ii), T = = 669 (approx)

# NCERT

4.26 The decomposition of hydrocarbon follows the equation $k = (4.5 × 1011s^{-1}) e^{-28000K/T}$ . Calculate $E_{a}$

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M manish

The Arrhenius equation is given by .................................(i) given equation, ............................(ii) by comparing equation (i) & (ii) we get, A= 4.51011 per sec Activation energy = 28000 (R = 8.314) = 232.792 KJ/mol

# NCERT

4.25 Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t _{1/2 } = 3.00$ hours. What fraction of sample of sucrose remains after 8 hours ?

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M manish

For first order reaction, given that half life = 3 hrs () Therefore k = 0.693/half-life = 0.231 per hour Now, = antilog (0.8024) = 6.3445 (approx) Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

# NCERT

4.24 Consider a certain reaction A $\rightarrow$ Products with $k = 2.0 \times 10 ^{-2 } s ^{-1}$ . Calculatethe concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol $L^{-1}$

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M manish

Given that, k = t = 100 s Here the unit of k is in per sec, it means it is a first-order reaction. therefore, Hence the concentration of rest test sample is 0.135 mol/L

# NCERT

4.23 The rate constant for the decomposition of hydrocarbons is $2.418 \times 10 ^{-5} s ^{-1}$ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

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M manish

Given that, k = = 179.9 KJ/mol T(temp) = 546K According to Arrhenius equation, taking log on both sides, = (0.3835 - 5) + 17.2082 = 12.5917 Thus A = antilog (12.5917) A = 3.9 per sec (approx)

# NCERT

4.22 The rate constant for the decomposition of N2O5 at various temperatures
is given below:

Draw a graph between ln k and 1/T and calculate the values of A and
$E_a$ . Predict the rate constant at 30° and 50°C.

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M manish

From the above data, T/ 0 20 40 60 80 T/K 273 293 313 333 353 () 3.66 3.41 3.19 3.0 2.83 0.0787 1.70 25.7 178 2140 -7.147 -4.075 -1.359 -0.577 3.063 Slope of line = According to Arrhenius equations, Slope = 12.30 8.314 = 102.27 Again, When T = 30 +273 = 303 K and 1/T =0.0033K k = When T = 50 + 273 = 323 K and 1/T = 3.1 K k = 0.607 per sec

# NCERT

4.21 The following data were obtained during the first order thermal decomposition of $SO_2 Cl_2$ at a constant volume.
$( SO_2Cl_2 g) \rightarrow SO_2 (g) + Cl_2 (g)$

Calculate the rate of the reaction when total pressure is 0.65 atm.

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M manish

The thermal decomposition of is shown here; After t time, the total pressure = So, thus, for first order reaction, now putting the values of pressures, when t = 100s when = 0.65 - 0.5 = 0.15 atm So, = 0.5 - 0.15 = 0.35 atm Thus, rate of reaction, when the...

# NCERT

4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

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M manish

Decompostion is represented by equation- After t time, the total pressure = So, thus, for first order reaction, now putting the values of pressures, when t =360sec when t = 270sec So,

# NCERT

4.19 A first order reaction takes 40 min for 30% decomposition. Calculate $t_{1/2}$

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M manish

For the first-order reaction, (30% already decomposed and remaining is 70%) therefore half life = 0.693/k = = 77.7 (approx)

# NCERT

4.18 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

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M manish

case 1- for 99% complition, CASE- II for 90% complition, Hence proved.

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