Question

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 1

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.