Question

The initial concentration of cane sugar in presence of acid was reduced from 0.20 to 0.10 M in 5 hours and to 0.05M in 10 hours. Value of K?
(A) 0.693
(B) 1.386
(C) 0.1386
(D) 3.465

Answer 1

Hint :To answer this question, we first need to understand what cane sugar is. Sugarcane, often known as sugar cane, is a type of long perennial grass in the genus Saccharum, tribe Andropogoneae, that is used to make sugar. The plants are 2-6 m tall, with strong, jointed, fibrous stalks that are high in sucrose, which accumulates in the internodes of the stalks.

Complete Step By Step Answer:
Rate of reaction - The rate of reaction, often known as the reaction rate, is the pace at which a reaction takes place. We quantify this statistically by measuring the rate at which reactants or products vanish or arise, where rate is defined as a time derivative. We'll focus on first-order reactions in this lesson. We can define the rate of these reactions as the rate of disappearance of this reactant because first-order reaction rates are only dependent on the concentration of one reactant.
The equilibrium constant - The equilibrium constant, K, expresses the connection between a reaction's products and reactants when it is in equilibrium with regard to a given unit. This article covers how to build equilibrium constant expressions and walks you through the calculations for both the concentration and partial pressure equilibrium constants.
$ {t_{1/2}}{\kern 1pt} {\kern 1pt} for{\kern 1pt} {\kern 1pt} first{\kern 1pt} {\kern 1pt} case{\kern 1pt} {\kern 1pt} is{\kern 1pt} {\kern 1pt} 5{\kern 1pt} hrs{\kern 1pt} {\kern 1pt} (as{\kern 1pt} {\kern 1pt} the{\kern 1pt} {\kern 1pt} concentration{\kern 1pt} {\kern 1pt} becomes{\kern 1pt} {\kern 1pt} half) \\
  {t_{1/2}}{\kern 1pt} {\kern 1pt} for{\kern 1pt} {\kern 1pt} \sec ond{\kern 1pt} {\kern 1pt} case{\kern 1pt} {\kern 1pt} (0.1{\kern 1pt} {\kern 1pt} to{\kern 1pt} {\kern 1pt} 0.05) \\
  reaction{\kern 1pt} {\kern 1pt} is{\kern 1pt} {\kern 1pt} first{\kern 1pt} {\kern 1pt} order \\
  K = \dfrac{{0.693}}{{{t_{1/2}}}} = \dfrac{{0.693}}{5} = 0.1386/hr $
So, the final answer is option (3) 0.1386.

Note :
Allowing a single reaction to reach equilibrium and then measuring the concentrations of each chemical involved in that reaction yields the numerical value of an equilibrium constant. The ratio between the product and reactant concentrations is computed. Because concentrations are recorded at equilibrium, the equilibrium constant remains constant regardless of initial concentrations for a specific reaction.