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In a chemical reaction, the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is
  • a)
    1
  • b)
    16
  • c)
    4
  • d)
    1/4
Correct answer is option 'C'. Can you explain this answer?
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In a chemical reaction,theinitial concentration of B was 1.5 times of ...
given
3 – 2x = 2 – x
⇒ x = 1
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The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.When C,H5OH and CH3COOH are mixed in equivalent proportion equilibrium is reached when 2/3 of acid and alcohol are used. How much ester will be present when 2 moles of acid were to react with 2 moles of alcohol? .

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumq.Above homogeneous reaction is carried out in a 2 litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If Kc for the reaction is then equilibrium concentration of C is

Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.Initial concentration of A is twice the initial concentration of `B. At equilibrium concentration of A and a are same then equilibrium constant for the reaction is

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In a chemical reaction,theinitial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction isa)1b)16c)4d)1/4Correct answer is option 'C'. Can you explain this answer?
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