
The rate constant for an isomerisation reaction is $4.5 \times {10^{ - 3}}$ /min. If the initial concentration of A is 1M, What will be the rate of reaction after 1 hour?
A. $3.44 \times {10^{ - 3}}mol{L^{ - 1}}{\min ^{ - 1}}$
B. $3.44 \times {10^3}$ $mol{L^{ - 1}}{\min ^{ - 1}}$
C. $1.86 \times {10^{ - 3}}$ $mol{L^{ - 1}}{\min ^{ - 1}}$
D. $1.86 \times {10^3}$ $mol{L^{ - 1}}{\min ^{ - 1}}$
Answer
523.5k+ views
Hint: The rate of the reaction depends on the concentrations of one or more than one reactants.
Complete step by step solution: the reaction is a first order reaction as it is an isomerisation reaction and its unit of rate constant is per min. Order of a reaction may be defined as the sum of powers or exponents to which the concentration terms are raised in the rate law expression.
Unit of rate constant=$\dfrac{{mol{L^{ - 1}}{{(time)}^{ - 1}}}}{{{{(mole{L^{ - 1}})}^n}}}$
Where n is the order of reaction. The differential rate expression for the first order reaction is
Rate=r=$
- \dfrac{{d[A]}}{{d[t]}} = k[A] \\
- \dfrac{{d[A]}}{{[A]}} = kdt \\
\int { - \dfrac{{d[A]}}{{[A]}} = k\int {dt} } \\
$
$
- \ln \dfrac{{{{[A]}_t}}}{{{{[A]}_0}}} = kt \\
kt = 2.303\log \dfrac{{[{A_ \circ }]}}{{[{A_t}]}} \\
$
T is 60 minute, [A] initial concentration and k=rate constant. Substituting the values we get
$
4.5 \times {10^{ - 3}} = \dfrac{{2.303}}{{60}}\log \dfrac{1}{{[{A_t}]}} \\
[{A_t}] = 0.7365mol{L^{ - 1}} \\
R = K[{A_t}] = 4.5 \times {10^{ - 3}} \times 0.7365 = 3.44 \times {10^{ - 3}}mol{L^{ - 1}}{\min ^{ - 1}} \\
$
Hence the correct option is option A.
Note: Order of a reaction may be zero, fractional or integer.
Complete step by step solution: the reaction is a first order reaction as it is an isomerisation reaction and its unit of rate constant is per min. Order of a reaction may be defined as the sum of powers or exponents to which the concentration terms are raised in the rate law expression.
Unit of rate constant=$\dfrac{{mol{L^{ - 1}}{{(time)}^{ - 1}}}}{{{{(mole{L^{ - 1}})}^n}}}$
Where n is the order of reaction. The differential rate expression for the first order reaction is
Rate=r=$
- \dfrac{{d[A]}}{{d[t]}} = k[A] \\
- \dfrac{{d[A]}}{{[A]}} = kdt \\
\int { - \dfrac{{d[A]}}{{[A]}} = k\int {dt} } \\
$
$
- \ln \dfrac{{{{[A]}_t}}}{{{{[A]}_0}}} = kt \\
kt = 2.303\log \dfrac{{[{A_ \circ }]}}{{[{A_t}]}} \\
$
T is 60 minute, [A] initial concentration and k=rate constant. Substituting the values we get
$
4.5 \times {10^{ - 3}} = \dfrac{{2.303}}{{60}}\log \dfrac{1}{{[{A_t}]}} \\
[{A_t}] = 0.7365mol{L^{ - 1}} \\
R = K[{A_t}] = 4.5 \times {10^{ - 3}} \times 0.7365 = 3.44 \times {10^{ - 3}}mol{L^{ - 1}}{\min ^{ - 1}} \\
$
Hence the correct option is option A.
Note: Order of a reaction may be zero, fractional or integer.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
