Read the passage given below and answer the questions that follow:The

1.	Read the passage given below and answer the questions that follow:The rate law for a chemical reaction relates the reaction rate with the concentrations or partial pressures of the reactants. For a general reaction aA + bB → C with no intermediate steps in its reaction mechanism, meaning that it is an elementary reaction, the rate law is given by r = k[A]x [B]y, where [A] and [B] express the concentrations of A and B in moles per litre. Exponents x and y vary for each reaction and are determined experimentally. The value of k varies with conditions that affect reaction rate, such as temperature, pressure, surface area, etc. The sum of these exponents is known as overall reaction order. A zero order reaction has a constant rate that is independent of the concentration of the reactants. A first order reaction depends on the concentration of only one reactant. A reaction is said to be second order when the overall order is two. Once we have determined the order of the reaction, we can go back and plug in one set of our initial values and solve for k.(i) Calculate the overall order of a reaction which has the following rate expression:Rate = k[A]1/2 [B]3/2(ii) What is the effect of temperature on rate of reaction ?(iii) What is meant by rate of a reaction ?(iv)(a) A first order reaction takes 77.78 minutes for 50% completion. Calculate the time required for 30% completion of this reaction. (log 10= 1, log 7 = 0.8450)OR(b) A first order reaction has a rate constant 1 x 10-3 per sec. How long will 5 g of this reactant take to reduce to 3g? (log 3 0.4771; log 5 = 0.6990)
Answer» (i) Rate = K [A]^1/2 [B]^3/2 Overall order of reaction = \(\frac12+\frac32=2\) (ii) According to Arrhenius equation , K = Ae^{-\((\frac{Eq}{RT})\)} As increase the temperature the value of exponential part become less negative and thus the value of K increasing. rate of reaction \(\propto\) K it means, increasing at the an reaction, generally rate of reaction increases. (iii) Rate of reaction is the speed at which reactant are converted into products. For an simple reaction - A \(\longrightarrow\) product rates of reaction = \(\frac{d[A]}{dt}\) or, rate of an reaction equal to the change in concentration of reactant per unit time. (iv) (a) t_1/2= 77.78 min \(\therefore K = \frac{0.693}{t_{1/2}}=\frac{0.693}{77.78}= 8.9 \times 10^{-3}min^{-1}\) \(\therefore t = \frac{2.303}{8.9\times10^{-3}}log\cfrac{a_0}{(a_0-\frac{a_0\times30}{100})}\) \(t = \frac{2.303}{8.9\times10^{-3}}log\frac{100}{70}\) \(t = \frac{2.303}{8.9\times10^{-3}}\times(log_{10}-log_7)\) \(t = \frac{2.303}{8.9\times 10^{-3}}(1-0.8450)\) t = 40.1 min (b) Rate constant K = 1 x 10^-3 s^-1 \(t = \frac{2.303}{10^{-3}}log\frac53\) t = 2.303 x 10³ (log 5 - log 3) t = 2.303 x 103 (0.6990 - 0.4771) t = 511.0 sec Hence, after 511 sec, 5g reactant reduced to 3g.

Read the passage given below and answer the questions that follow:The rate law for a chemical reaction relates the reaction rate with the concentrations or partial pressures of the reactants. For a general reaction aA + bB → C with no intermediate steps in its reaction mechanism, meaning that it is an elementary reaction, the rate law is given by r = k[A]x [B]y, where [A] and [B] express the concentrations of A and B in moles per litre. Exponents x and y vary for each reaction and are determined experimentally. The value of k varies with conditions that affect reaction rate, such as temperature, pressure, surface area, etc. The sum of these exponents is known as overall reaction order. A zero order reaction has a constant rate that is independent of the concentration of the reactants. A first order reaction depends on the concentration of only one reactant. A reaction is said to be second order when the overall order is two. Once we have determined the order of the reaction, we can go back and plug in one set of our initial values and solve for k.(i) Calculate the overall order of a reaction which has the following rate expression:Rate = k[A]1/2 [B]3/2(ii) What is the effect of temperature on rate of reaction ?(iii) What is meant by rate of a reaction ?(iv)(a) A first order reaction takes 77.78 minutes for 50% completion. Calculate the time required for 30% completion of this reaction. (log 10= 1, log 7 = 0.8450)OR(b) A first order reaction has a rate constant 1 x 10-3 per sec. How long will 5 g of this reactant take to reduce to 3g? (log 3 0.4771; log 5 = 0.6990)

Answer»

(i) Rate = K [A]^1/2 [B]^3/2

Overall order of reaction = \(\frac12+\frac32=2\)

(ii) According to Arrhenius equation ,

K = Ae^{-\((\frac{Eq}{RT})\)}

As increase the temperature the value of exponential part become less negative and thus the value of K increasing.

rate of reaction \(\propto\) K

it means, increasing at the an reaction, generally rate of reaction increases.

(iii) Rate of reaction is the speed at which reactant are converted into products.

For an simple reaction -

A \(\longrightarrow\) product

rates of reaction = \(\frac{d[A]}{dt}\)

or, rate of an reaction equal to the change in concentration of reactant per unit time.

(iv)

(a) t_1/2= 77.78 min

\(\therefore K = \frac{0.693}{t_{1/2}}=\frac{0.693}{77.78}= 8.9 \times 10^{-3}min^{-1}\)

\(\therefore t = \frac{2.303}{8.9\times10^{-3}}log\cfrac{a_0}{(a_0-\frac{a_0\times30}{100})}\)

\(t = \frac{2.303}{8.9\times10^{-3}}log\frac{100}{70}\)

\(t = \frac{2.303}{8.9\times10^{-3}}\times(log_{10}-log_7)\)

\(t = \frac{2.303}{8.9\times 10^{-3}}(1-0.8450)\)

t = 40.1 min

(b) Rate constant K = 1 x 10^-3 s^-1

\(t = \frac{2.303}{10^{-3}}log\frac53\)

t = 2.303 x 10³ (log 5 - log 3)

t = 2.303 x 103 (0.6990 - 0.4771)

t = 511.0 sec

Hence, after 511 sec, 5g reactant reduced to 3g.

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