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PCl5(g) → PCl3(g) + Cl2(g)
In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)
[Given log 5 = 0.6989]
    Correct answer is '1'. Can you explain this answer?
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    PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the c...
    PCl5(g) refers to the chemical compound phosphorus pentachloride in its gaseous state. PCl5 is a covalent compound that consists of one phosphorus atom bonded with five chlorine atoms. In its gaseous state, PCl5 forms a white fuming gas with a pungent odor. It is highly reactive and is commonly used as a chlorinating agent in various chemical reactions.
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    PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the c...

    Or x = 1 (Nearest integer)
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    There are many reactions which obey a first order rate equation although it reality they are bi- or ter-molecular. As an example of these may be taken the decomposition of Carbonyl sulfide in water, namely, COS + H20 002 + H2SAccording to the law of mass action this reaction should be second order with the rate dependent on the concentration of both the carbonyl sulfide and the water. Actually however, the rate is found to be first order with respect to the carbonyl sulfide and independent of the water Reactions exhibiting such behaviour are said to be pseudo-molecular.The pseudo-unimoecuar nature of this reaction is explainable by the fact that water is present in such excess that its concentration remains practically constant during the course of the reaction. Under these condition b x = b, and the rate equation becomesOn integration this leads towhich is the equation for a first order reaction. It is evident, however, that the now constant k is not independent of the concentration, as is the case with true first order constants, but may vary with b if the latter is changed appreciably, When such is the case, the true constant k2 can be obtained from k by dividing the latter by b. pseudo-molecular reactions are encountered whenever one or more of the reactants remain constants during the course of an experiment. This is the case with reactions conducted in solvents which are themselves one of the reactants, as in the decomposition of carbonyl sulfide in water, or in the esterification of acetic anhydride in alcohol(CH3C0)20 + 2C2H5OH 2CH3C00C2H5 + H20Again, this is also true of reactions subject to catalysis, in which case the concentration of the catalyst does not change. The decomposition of diacetone alcohol to acetone in aqueous solution is catalysed by hydroxyl ions, with the rate proportional to the concentration of the alcohol and that of the base. Since the concentration of the base does not change within any one experiment, however, the rate equation reduces to one of first order with respect to the alcohol. But the rateconstant k obtained for various concentrations of base are not identical, as may be seen from table. To obtain from these the true second order velocity constant, the ks must be divided by the hydroxyl ion concentration. When this is done excellent k2 values result, as column 3 indicatesTable : Decomposition of diacetone alcohol in water at 25C (Catalyst : NaOH)Q.By what factor does the rate of reaction of diacetone alcohol in water solution change if p0H is increased by 2 units other things remaining same ?

    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

    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)At what time is the rate of change of concentration of A equal to rate of change of concentration of C in magnitude ?

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    PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer?
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    PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer?.
    Solutions for PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
    Here you can find the meaning of PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer?, a detailed solution for PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? has been provided alongside types of PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice PCl5(g) →PCl3(g) + Cl2(g)In the above first order reaction, the concentration of PCl5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is X × 10-2 min-1. The value of x is ___________.(Nearest integer)[Given log 5 = 0.6989]Correct answer is '1'. Can you explain this answer? tests, examples and also practice JEE tests.
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