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Chemistry Test - 4
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Chemistry Test - 4
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  • Question 1/10
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    Directions: In the following question, a statement of assertion is given, followed by a corresponding statement of the reason below it. Choose the correct option.

     

    Assertion: Cu2+ and Cd2+ are separated by first adding KCN solution and then, passing H2S gas.

     

    Reason: KCN reduces Cu2+ to Cu+ and forms a complex with it.

    Solutions

    Both assertion and reason are true, but the reason is not a correct explanation for the assertion.

    Cu+2 makes a more stable complex with CN- which is soluble and therefore the solution, now contains, the complex Cu+2 ion and the free Cd+2 ion. Hence only the sulphide of Cadmium will precipitate out on the passage of H2S gas.

  • Question 2/10
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    Directions: In the following question, a statement of Assertion is given; followed by a corresponding statement of Reason just below it. Out of the statements mark the correct answer as:

     

    (a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

     

    (b) If both assertion and reason are true, but the reason is not the correct explanation of the assertion.

    (c) If assertion is true statement, but reason is false.

    (d) If both assertion and reason are false.

     

    Assertion: According to transition state theory, for the formation of an activated complex; one of the vibrational degree of freedom is converted into a translational degree of freedom.

     

    Reason: Energy of the activated complex is higher than the energy of reactant molecules.

    Solutions

    Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. TST is used primarily to understand qualitatively how chemical reactions take place. In other words, one of the vibrational degrees of freedom is converted into translational degree of freedom.

     

    An activated complex acts as an intermediary between the reactants and the products of the reaction. The energy of the activated complex is higher than that of reactants or the products, and the state is temporary. If there is not sufficient energy to sustain the chemical reaction, the activated complex can reform into the reactants in a backward reaction. With proper energy, though, the activated complex forms the products in a forward reaction.

     

  • Question 3/10
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    For a hypothetical reaction 2A + 3B→2C, the rate law expression is = dx/dt k[A] [B]2. The order of the reaction is

    Solutions

    In chemical kinetics, the order of reaction with respect to a given substance (such as reactant, catalyst or product) is defined as the index, or exponent, to which its concentration term in the rate equation is raised. For the given expression, according to rate law, order of reaction is: 2 + 1 = 3.

  • Question 4/10
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    The plot of ln C0 /Ct against t is a straight line; the order of the reaction is expected to be:

    Solutions

  • Question 5/10
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    A reaction was found to be in second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled with everything else kept the same, the rate of reaction will

    Solutions

  • Question 6/10
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    Unit of rate constant of zero order reaction is

    Solutions

    According to rate law for zero order reaction,

    Rate = K [A]0

    = K = Mol L-1s-1

    Thus, unit of rate constant is: Mol L-1s-1

  • Question 7/10
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    The gas-phase reaction of nitric oxide and bromine yields nitrosyl bromide as 2NO(g) + Br2(g) →2NOBr(g). What is the order of the reaction?

    Solutions

    Rate law is ,

    rate=k[NO]2[Br2]

    hence, wrt NO, order=2

    wrt Br2, order=1

    ∴overall order = 2+1=3

  • Question 8/10
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    For As2S3 sol, the maximum coagulation power is of

    Solutions

    As2S3 is a negatively charged sol.

    Hence, the cation of the electrolyte will be the active ion and option (4) is correct.

  • Question 9/10
    1 / -0

    Bredig's method can not be used for the preparation of collodial sol of

    Solutions

    Bredig's method cannot be used for the preparation of colloidal sol of sodium as it does not give satisfactory results.

  • Question 10/10
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    If C6H6(I) + 15/2 O2(g)→3H2O(I) + 6CO2(g); ΔH= – 3264.6 kJ mol–1, then the energy obtained by burning 3.9 g of benzene in the air is

     

    Solutions

    78 g of C6H6 gives heat

    = 3264.6 kJ

    ∴3.9 g of C6H6 will give heat

    = (3264.6 / 78) x 3.9= 163.23 kJ

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