
A first order reaction has a rate constant $1.15 \times {10^{ - 3}}{s^{ - 1}}$. How long will 5 g of this reactant take to reduce to 3 g?
(A) 444 s
(B) 400 s
(C) 528 s
(D) 669 s
Answer
572.7k+ views
Hint: The reaction is called first order reaction when Rate of reaction depends upon concentration of one reactant only.
Example:
$2{H_2}{O_{2(g)}} \to 2{H_2}{O_{(l)}} + {O_2}$
Rate$ = k[{H_2}{O_2}]$
This reaction is first order because Rate of reaction is the first power of ${H_2}{O_2}$ concentration.
Formula used: $t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{A_o}}}{{{A_t}}}$
Complete step by step answer:
As the reaction is first order reaction therefore formula used is
$t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{A_o}}}{{{A_t}}}$______(1)
Where,
t = time
k = Rate constant of first order reaction
${A_o}$ = Initial concentration
${A_t}$ = Final concentration
The values are
$k = 1.15 \times {10^{ - 3}}{s^{ - 1}}$
${A_o} = 5gm$
${A_t} = 3gm$
$t = ?$
Substituting this values in equation number (1) we get
$t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}}\log \dfrac{5}{3}$
$t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}}\log 1.66$
Value of $\log 1.66 = 0.22$
Putting this value we have $t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}} \times \dfrac{{0.22}}{1}$
$t = \dfrac{{2.303 \times 0.22 \times {{10}^3}}}{{1.55}}$
$t = \dfrac{{506.66}}{{1.55}}$
$t = 440.57\sec .$
Therefore, from the above explanation the correct option is (A) 444 s.
Additional information: Overall order of reaction is the sum of the exponent to which concentration terms in the rate law are raised.
Let us consider a reaction.
$aA + bB \to cA + dD$
Here the number of molecules of A reacts with the B molecule.
The order of reaction is not related to the stoichiometric equation of reaction.
It means order of reaction is not equal to = a + b
It is an experimentally determined quantity.
Note: The reaction order is always defined in terms of the concentration of reactant and not of product.
Rate constant for first order depends only on time not concentration reactant.
Example:
$2{H_2}{O_{2(g)}} \to 2{H_2}{O_{(l)}} + {O_2}$
Rate$ = k[{H_2}{O_2}]$
This reaction is first order because Rate of reaction is the first power of ${H_2}{O_2}$ concentration.
Formula used: $t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{A_o}}}{{{A_t}}}$
Complete step by step answer:
As the reaction is first order reaction therefore formula used is
$t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{A_o}}}{{{A_t}}}$______(1)
Where,
t = time
k = Rate constant of first order reaction
${A_o}$ = Initial concentration
${A_t}$ = Final concentration
The values are
$k = 1.15 \times {10^{ - 3}}{s^{ - 1}}$
${A_o} = 5gm$
${A_t} = 3gm$
$t = ?$
Substituting this values in equation number (1) we get
$t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}}\log \dfrac{5}{3}$
$t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}}\log 1.66$
Value of $\log 1.66 = 0.22$
Putting this value we have $t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}} \times \dfrac{{0.22}}{1}$
$t = \dfrac{{2.303 \times 0.22 \times {{10}^3}}}{{1.55}}$
$t = \dfrac{{506.66}}{{1.55}}$
$t = 440.57\sec .$
Therefore, from the above explanation the correct option is (A) 444 s.
Additional information: Overall order of reaction is the sum of the exponent to which concentration terms in the rate law are raised.
Let us consider a reaction.
$aA + bB \to cA + dD$
Here the number of molecules of A reacts with the B molecule.
The order of reaction is not related to the stoichiometric equation of reaction.
It means order of reaction is not equal to = a + b
It is an experimentally determined quantity.
Note: The reaction order is always defined in terms of the concentration of reactant and not of product.
Rate constant for first order depends only on time not concentration reactant.
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