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Chemical Kinetics Test - 68

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Chemical Kinetics Test - 68
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  • Question 1
    1 / -0
    For a chemical reaction to occur, all of the following must happen except:
    Solution
    For a chemical reaction to occur, bonds must break in reactant so that new bonds are formed. Moreover, reactants must collide with energy greater than activation energy for change to occur. Also, reactant must have proper orientation for reaction.

    However, it is not necessary that enough number of collisions must occur.
  • Question 2
    1 / -0
    Consider figure and mark the correct option.

    Solution
    The product is less stable since the forward reaction is not favorable. The reaction is endothermic ($$\Delta H>0$$). 

    Option A is correct.

  • Question 3
    1 / -0
    For reaction $$A \rightarrow B$$, the rate law expression is $$-\dfrac{d[A]}{dT}=k[A]^{1/2}$$. If initial concentration of $$[A]$$ is $$[A]_0$$ then, which of the following statement is incorrect?

    Solution
    From the rate law,

    $$R=K[A]^{1/2}$$

    Thus order of Reaction is, $$n=1/2$$

    General formula for $$t_{1/2}$$

    $$t_{1/2}=\cfrac {2^{n-1}-1}{K[A_o]^{n-1}}$$ or $$t_{1/2} \propto \cfrac {1}{[A]^{n-1}}$$

    For, $$n= 1/2$$

    $$t_{1/2} \propto \cfrac {1}{[A_o]^{1/2-1}} \Rightarrow t_{1/2} \propto \cfrac {1}{[A_o]^{-1/2}}$$

    or, $$t_{1/2} \propto [A_o]^{1/2}$$

    Thus, statement C is incorrect.
  • Question 4
    1 / -0
    Rate of formation of $${ SO }_{ 3 }$$ according to the reaction $$2{ SO }_{ 2 }+O_{ 2 }\rightarrow 2{ SO }_{ 3 }\quad is\quad 1.6\times { 10 }^{ -3 }kg\quad min^{ -1 }$$. Hence rate at which $$SO_{ 2 }$$ reacts is:
    Solution
    $$2 SO_2 + O_2 \rightleftharpoons  2 SO_3$$

    $$\dfrac{1}{2}\dfrac{d[SO_3]}{dt} = 1.6 \times 10^{-3} kg/min = \dfrac{1.6 \times 10^{-3}}{80} mol/min$$                             $$[\because M_{SO_3}= 80 \times 10^{-3} Kg/mol]$$

    $$\dfrac{-1}{2}\dfrac{d[SO_2]}{dt} = \dfrac{+1}{2}\dfrac{d[SO_3]}{dt}$$

    $$= -\dfrac{1.6 \times 10^{-3}}{80 \times 10} $$ mol/min

    $$= -\dfrac{1}{50} \times 64 \times 10^{-3} = -1.28 \times 10^{-3} $$ kg/min
  • Question 5
    1 / -0

    Directions For Questions

    A group of mountain climbers set up a camp at 3 km altitude and experience a barometric pressure of $$0.69atm$$. They discover that pure water boils at $$90^oC$$ and it takes $$300min$$ of cooking to make a "three-minute" egg.

    ...view full instructions

    What is the activation energy for this change?
    Solution

  • Question 6
    1 / -0
    Which of the following reaction has zero activation energy?
    Solution
    Among the given reactions
    $${ \dot { CH }  }_{ 3 }+{ \dot { CH }  }_{ 3 }\rightarrow { CH }_{ 3 }-{ CH }_{ 3 }$$ has zero activation energy.
    All the other reactions are following a free radial which needs energy in the form $$U.V$$ light which is high activation energy.
    Where as in the above reaction the free radicals are bonded together by lowering their energy thus by increasing the stability.
    So, the reaction 
    $${ \dot { CH }  }_{ 3 }+{ \dot { CH }  }_{ 3 }\rightarrow { CH }_{ 3 }-{ CH }_{ 3 }$$ has zero activation energy.
  • Question 7
    1 / -0
    The activation energy of a reaction is $$58.3\ kJ/mole$$. The ratio of the rate constants at $$305K$$ and $$300K$$ is about:
    [$$R=8.3\ J{k}^{-1}{mol}^{-1}$$ and Antilog $$0.1667=1.468$$]
    Solution
    $$log\quad \frac { { K }_{ 2 } }{ { K }_{ 1 } } =\frac { { E }_{ o } }{ 2.303\quad X\quad R } \quad \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } }  \right] $$

    $$=\quad \frac { 58.3 }{ 2.303\quad X\quad 8.3 } \left[ \frac { 305-300 }{ 305\quad X\quad 300 }  \right] $$

    $$=\quad 3.0499\quad X\quad 5.4644\quad X\quad 1{ 0 }^{ -5 }$$

    $$=1.666\quad X\quad 1{ 0 }^{ -4 }$$

    =$$log\quad \frac { { K }_{ 2 } }{ { K }_{ 1 } } =0.1666\quad X\quad 1{ 0 }^{ -3 }$$

    =$$\frac { { K }_{ 2 } }{ { K }_{ 1 } } =antilog(0.1666)$$

    $$=1.468\approx 1.5$$

    Hence, 1.5 is the answer.
  • Question 8
    1 / -0
    A reaction takes place in three steps with the rate constant $$k_1, k_2$$ and $$k_3$$. The overall rate constant $$k=\dfrac{k_1(k_2)^{1/2}}{k_3}$$. If activation energies are $$40, 30$$ and $$20\ kJ$$ for step $$I, II$$ and $$III$$ respectively, the overall activation energy of reaction will be:
    Solution
    Over all activation energy

    $$=E_{1}+\dfrac{E_{2}}{2}-E_{3}$$

    $$=40+\dfrac{30}{2}-20$$

    $$=35 KJ$$
  • Question 9
    1 / -0
    Which of the following reactants can be used to distinguish between benzaldehyde and Benzyl alcohol?
    Solution

  • Question 10
    1 / -0
    It is often stated that, near room temperature, a reaction rate doubles if the temperature increases by $${ 10 }^{ \circ  }C$$ Calculate the activation energy of a reaction that obeys this rule exactly. 
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