Question:

For a first order reaction at 27$^{\circ}$C, the ratio of time required for 75% completion to 25% completion of reaction is

Updated On: May 20, 2024
  • 3
  • 2.303
  • 4.8
  • 0.477
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The Correct Option is C

Solution and Explanation

For a first order reaction,
$\hspace25mm t =\frac{2.303}{k} \, log_{10} \, \frac{a}{a-x}$
Let initial amount of reactant is 100.
$\hspace20mm \frac{t_1}{t_2} = \frac{log \frac{100}{100-75}}{log \frac{100}{100-25}} \, [\therefore \, k \, remains \, constant]$
$\hspace20mm =\frac{log \frac{100}{25}}{log \frac{100}{75}} = \frac{log \, 4}{log \, 4/3}$
$\hspace20mm =\frac{log \, 4}{log \, 4 - log \, 3}$
$\hspace20mm =\frac{2 \times 0.3010}{2 \times 0.3010-0.4771}$
$\hspace20mm = \frac{0.6020}{0.1249} = 4.81$
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Factors Affecting The Reaction Rate:

  • The concentration of Reactants - According to collision theory, which is discussed later, reactant molecules collide with each other to form products.
  • Nature of the Reactants - The reaction rate also depends on the types of substances that are reacting.
  • Physical State of Reactants - The physical state of a reactant whether it is solid, liquid, or gas can greatly affect the rate of change.
  • Surface Area of Reactants - When two or more reactants are in the same phase of fluid, their particles collide more often than when either or both are in the solid phase or when they are in a heterogeneous mixture. In a heterogeneous medium, the collision between the particles occurs at an interface between phases. Compared to the homogeneous case, the number of collisions between reactants per unit time is significantly reduced, and so is the reaction rate.
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