Chemistry Test - 4

The correct answer is option 3.

Key Points

• A decomposition reaction is a type of chemical reaction in which a compound breaks down into two or more substances.
• The starting substance is called reactant and the end substances are called products.

​

• Example: Water splits into Hydrogen and Oxygen upon electrical energy.

​

Additional Information

• Decomposition reaction generally takes energy to break the existing bonds in the reactant.
• Based on the type of energy the reactant will be decomposed thermally, electrically, photochemically or in presence of a catalyst.
• It can be observed practically in many common activities like:​
  1. ​Digestion of food (complex food breaking down into nutrient materials)
  2. Fizz effect in many bottled drinks (CO2 gas bubbles from Carbonic acid)
  3. Baking a cake (SodiumBicarbonate into Sodium Hydroxide and Carbon Dioxide)​

Concept:

Rate of a Reaction:

• The rate of a reaction is the velocity of a reaction.
• It is the amount of chemical change occurring with time.
• As the reaction continues, the amount or concentration of reactants decreases, and the concentration of products increases.

Thus it can be regarded as the rate of decrease of reactants or the rate of increase of products.

Rate Law:

• Rate law states that the rate of a reaction is proportional to the concentration of the reactants raised to the power of the order of the reaction.

In mathematical terms, we can say that

Rate = - dc/dt ; the rate of decrease of reactants

Rate = - dx/dt; the rate of increase of reactants

→ Rate = k [C]; 

where 'k' = rate constant and 'n' = order of a reaction.

Calculation:
Given:

• The reaction is 2NO + Br2 → 2NOBr.
• The steps of the reaction are:

\(\rm NO \ + \ Br_2 \ \xrightarrow{Fast} \ NOBr_2\)  -- (I)

\(\rm NOBr_2 \ + \ NO \ \xrightarrow{Slow} \ 2NOBr\) --- (II)

The rate-determining step is \(\rm NOBr_2 \ + \ NO \ \xrightarrow{Slow} \ 2NOBr\) the because it is the slowest step.

According to rate law, the rate = k [NOBr₂][NO] - (a)

But the NOBr₂ is an intermediate and thus its concentration should be replaced.

For Reaction (I)

→ We know Equilibrium Constant K_c is 

→ \(K_c = \frac{[NoBr_2]}{[NO][Br_2]}\)  → [NOBr₂] = K_c × [NO] × [Br₂] --- (b)

Substituting (b) in Equation (a) we get 

Rate = k × Kc × [NO] × [Br2] × [NO]

∴ Rate = k' × [NO]² × [Br2]

Hence, for the reaction 2NO + Br2 → 2NOBr, the rate law is k [Br2][NO]².

Additional Information The rate-determining step of a reaction:

• There are multiple steps involved in a reaction involving intermediates.
• Each and every step has a different rate and it affects the overall rate of the reaction.
• Faster steps have a low effect on the reaction rate but when a step rate is slow, it slows down the whole reaction.
• Hence, the slowest step of the reaction becomes the rate-determining step.
• This is also known as the Bottleneck principle, as the rate of flow of water through the bottle depends on the size of the neck of the bottle.

Explanation:

⇒ The order of reaction with respect to a given substance (reactant, catalyst or product) is defined as the index or exponent to which its concentration term in the rate equation is raised.

For typical rate equation of form

r = k[A]x [B]y

⇒ Rate = k [CHCL3] [Cl2]1/2

⇒ Compare from the typical rate equation and we will get: x = 1, y = 1/2.

⇒ Overall reaction order = x + y

Here, x = 1, y = 1/2

∴ order of reaction = 1.5

Concept:

Rate Law:

• Rate law states that the rate of a reaction is proportional to the concentration of the reactants raised to the power of the order of the reaction.

In mathematical terms, we can say that

\( Rate = - (dc/dt) \); the rate of decrease of reactants

\(= dx/dt\) ; the rate of increase of reactants

\(rate = k[C]^n\) ; where 'k' = rate constant and 'n' = order of a reaction.

​Order of a reaction:

• The order of a reaction is the number of concentration terms on which the reaction rate depends.
• If multiple concentration terms are involved, then the order of a reaction is the sum of all powers.
• For example if -

\(- (dc/dt) = k[A]^x[B]^y ,\) then

\( order = x + y​\)

​The rate law for the nth-order reaction is:

 \(\left( {n - 1} \right)kt = \frac{1}{{{{\left[ A \right]}^{n - 1}}}} - \frac{1}{{{{\left[ {A^\circ } \right]}^{n - 1}}^{}}}\)

\(\left( {n - 1} \right)k = {1\over t} \frac{1}{{{{\left[ A \right]}^{n - 1}}}} - \frac{1}{{{{\left[ {A^\circ } \right]}^{n - 1}}^{}}}\)

\(unit\;of\;k = {L^{n - 1\;}}\;{Mol^{1 - n}}\;{t^{ - 1}}\)

Calculation:

Given:

The value of rate constant for the reaction 2NO2 + F2 → 2NO2F is = 38 dm3 mol^-1 s-1 

The temperature of the reaction = 300K

• We know that the unit of the rate constant is given by:

\(unit\;of\;k = {L^{n - 1\;}}\;{Mol^{1 - n}}\;{t^{ - 1}}\)

• Comparing the given unit of 'k' with the above equation is:

Mol^1-n = Mol^-1

• Comparing the powers as the bases are equal, we get:

1 - n = - 1

or, n = 1 + 1 = 2.

Hence, the order of the reaction is 2.

CONCEPT:

• Half-life: The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration. It is denoted by the symbol ‘t1/2’ and is usually expressed in seconds.
• The half life period of first order reaction may be calculated as given below:

​N = N0 / 2n

Where, n = no. of half-lives

n  = Total time (t) / Half-life period (t½)

N0 = Initial concentration of the substance           

N = Amount of radioactive substance left after n- half-live

Calculations:

Given:  t½ = 20 min ; Let N0 = x ;

After completion 75% remaining amount of reactant

N = 25% of x = x / 4

To find: t =?

We know, N = N0 / 2n

 ⇒ x / 4 = x / 2n

 ⇒ 2n = 4   (·.· 22 = 4)

 ⇒  n = 2 

Also, n = Total time (t) / Half-life period (t½)

⇒ 2 = t / 20

⇒ t = 40 min

Hence, first order reaction takes 40 minutes to 75% completion.

Concept:

The turnover frequency quantifies the specific activity of a catalytic centre for a given reaction occurring at specific conditions by the number of molecular catalytic reaction occurring at the centre per unit time.

The formula is:

The volume rate of reaction / Number of centres per volume

= Rate of reaction / Concentration of catalyst

=\({Volume× mole \over moles× time × volume} = time^{-1}\)

For most relevant industrial application reactions, the TOF is between 10^-2 to 10² s^-1.

For enzyme catalysis, it is between 10³ to 10⁷ s-1.

Calculation:

Given:

The 1.0 μM of a catalyst yields 1.0 mM of a product.

Time taken for product formation = 10 secs.

The amount of product formed = 1.0 mM = 1 × 10^-3

The amount of catalyst used 'Q'= 1.0μM = 1.0 × 10^-6

The rate of reaction is:

\(r = {1\times 10^{-3}\over 10}=1\times 10^{-4}\)

The turnover frequency TOF =

\({r\over Q} = {1\times 10^{-4}\over 1 \times 10^{-6}}={1\times 10^2}s^{-1}\)

Hence, the turnover frequency (TOF) of the reaction (s-1) is 102.

Explanation:- 

First Order Reaction is a chemical reaction in which the rate of the reaction is directly proportional to the concentration of one reactant.

R → P, ln [A] = -k t + ln [A0]

Calculation:- 

⇒ [R]t = (k2 / k1 + k2) × [P]o (1 - e-1(k1+k2)t) 

⇒ [R]t =  (15 × 10-2 min-1 / (5 + 15) × 10-2 min-1)  [4 mol L-1] (1 - e-1(5+15)x 10-2 x 10) 

⇒ [R]t = 3 (1 - e-20 x 1/100 x 10) 

⇒  [R]t = 2.59 mol L-1

Concept:

Rate of a Reaction:

• The rate of a reaction is the velocity of a reaction.
• It is the amount of chemical change occurring with time.
• As the reaction continues, the amount or concentration of reactants decreases, and the concentration of products increases.

Thus it can be regarded as the rate of decrease of reactants or the rate of increase of products.

The relationship between reaction rate and temperature:

• According to Arrhenius, the molecules in a reaction must possess a certain amount of energy in order to undergo a chemical reaction.
• This means that before the reaction occurs, the energy of the molecules must be raised in order to bring any transformation.
• This energy-rich state is called the activated state of the molecules.
• The extra energy required to attain this minimum energy has to be supplied externally and is known as the activation energy
• Reactions with low requirements of activation energy have faster rates.
• The higher the activation energy, the slower is the reaction rate.
• The relation between activation energy and temperature and the rate of reaction is given by the Arrhenius equation:

The Arrhenius equation is,

\(k = A{e^{ - {E_a}/RT}}\)

Where, k = rate constant A = Arrhenius constant Ea = activation energy T = temperature in K

Explanation:

• Taking logarithm on both sides of Arrhenius equation, we get:

ln k = ln A - \(\frac{{{E_a}}}{{RT}}\).

• The plot of lnk vs 1/RT looks like:

• The plot is a straight line.
• The equation of a straight line is y = mx + c, where m = slope and c = intercept.
• Comparing it with the Arrhenius Equation 

ln k = ln A - \(\frac{{{E_a}}}{{RT}}\).,

• We get the slope

m = -Ea / R and intercept = lnA.

Hence, the slope of the plot lnK vs 1/RT is -Ea / R.

Additional Information

• It is seen that for every 10°C rises of temperature, the rate of reaction doubles.

Concept:

Rate of the reaction

• The speed of a chemical reaction is called the rate of the reaction.
• As the reaction proceeds, the amount of reactants decreases whereas the amount of products increases. 

Consider a chemical reaction, 

aA + bB → products. 

The rate of the reaction is directly proportional to the concentration of the reactants.

Rate = K[A]^a [B]^b

K-rate constant.

[A]-Concentration of reactant A.

[B]- Concentration of reactant B.

The most important orders of reaction 

• Zero-order reaction.

Rate of the reaction is independent of the concentration of the reactants.

Rate =K

• First-order reaction.

Rate of the reaction depends on the concentration of one of the reactants involved in the reaction. 

Rate=K[A]

• Second-order reaction.

Rate of the reaction depends upon the concentration of two reactant molecules. 

Rate= K[A]²

Explanation:

The reaction in which rate doesn't depend on the concentration of reacting species is known as Zero order reaction. 

Consider a zero-order reaction, 

A → products

Rate = K[A]⁰

Rate = K

So, If the rate of reaction does not depend upon the initial concentration of reactant, the order of the reaction is Zero. 

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back. Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

According to equilibrium constant, K_c

\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{product}}} \right]}^{\rm{m}}}}}{{{{\left[ {{\rm{reactant}}} \right]}^{\rm{n}}}}}\)

Consider the given first equation:

\({A_2}\left( g \right)\; + \;{B_2}\;\left( g \right)\begin{array}{*{20}{c}} {{K_1}}\\ \rightleftharpoons \end{array}\;2AB\left( g \right)\)

\({K_1} = \frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}\)

Consider the given second equation:

\(6AB\left( g \right)\begin{array}{*{20}{c}} {{K_2}}\\ \rightleftharpoons \end{array}\;3{A_2}\left( g \right) + 3{B_2}\left( g \right)\)

\({K_2} = \frac{{{{\left[ {{A_2}} \right]}^3}{{\left[ {{B_2}} \right]}^3}}}{{{{\left[ {AB} \right]}^6}}}\)

\(= \frac{1}{{{{\left( {\frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}} \right)}^3}}}\)

\(= \frac{1}{{K_1^3}}\)

 → K2 = K1-3