CBSE 12 Class Exams Questions and Answers

NCERT

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.

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

Discuss the continuity of the function f where f is defined by 2x if x<0, 0 if 0<=x<=1, 4x if x>1

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...

NCERT

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.

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

NCERT

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 × 10 ^ 10 s ^ –1. Calculate k at 318K and Ea.

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

The decomposition of A into product has value of k as 4.5 × 10 ^ 3 s ^ –1 at 10°C and energy of activation 60 kJ mol ^ –1. At what temperature would k be 1.5 × 10 ^ 4s ^–1?

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

The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 10 ^ 4K/T

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

The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s ^–1) e^-28000K/T Calculate E

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

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

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

Consider a certain reaction A ® Products with k = 2.0 × 10^–2 s^–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L ^–1.

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

The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s– 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

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

The rate constant for the decomposition of N_2 O_5 at various temperatures is given below:

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

The following data were obtained during the first order thermal decomposition of SO_2 Cl_2 at a constant volume. ( SO_2 Cl_2 g) gives SO_2 (g) plus Cl_2 (g)

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

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

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

A first order reaction takes 40 min for 30% decomposition. Calculate t 1/2.

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

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

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.

NCERT

During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

4.17 During nuclear explosion, one of the products is $^{90} Sr$ with half-life of 28.1 years. If $1 \mu g$ of $^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

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M manish

Given, half life = 21.8 years = 0.693/21.8 and, by putting the value we get, taking antilog on both sides, [R] = antilog(-0.1071) = 0.781 Thus 0.781 of will remain after given 10 years of time. Again, Thus 0.2278 of will remain after 60 years.

NCERT

The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

4.16 The rate constant for a first order reaction is $60 s^{-1}$ . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?