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Chemical Kinetics Test - 19

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Chemical Kinetics Test - 19
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  • Question 1
    1 / -0
    A substance having initial concentration (a) reacts according to zero order kinetics. How much time will the reaction take to reach the completion?
    Solution
    For a zero order reaction, [A0][A]=Kt[A_{0}] - [A] = Kt

    When the reaction is complete, [A]=0[A] =0.

    [A0][A]=[A0]0=[A0]=Ktcompletion[A_{0}] - [A] =[A_{0}] -0=[A_{0}] = Kt_{completion}

    tcompletion=[A0]K=aKt_{completion}=\dfrac {[A_{0}] } {K}=\dfrac{a}{K}
  • Question 2
    1 / -0
    If the initial concentration is reduced to 1/4th in a zero order reaction, then the time taken for half the reaction to complete:
    Solution
    For a zero order reaction, t1/2=[A]02kt_{1/2}=\dfrac{[A]_0}{2k}

    Let [A]0=a[A]_0=a

    t1/2=a2kt_{1/2}=\dfrac{a}{2k}   ......(1)

    When the intial concentration is reduced to one fourth,  [A]0=14×a.[A]_0=\dfrac {1} {4} \times a.

    t1/2=14×a2kt'_{1/2}=\dfrac{\dfrac {1} {4} \times a}{2k}   ......(2)

    By dividing equation (2) by (1), we get
    t1/2t1/2=14×a2ka2k=14\dfrac {t'_{1/2}} {t_{1/2}} =\dfrac {\dfrac{\dfrac {1} {4} \times a}{2k}} {\dfrac{a}{2k}} =\dfrac {1} {4}

    Hence, option C is correct.
  • Question 3
    1 / -0
    At 400K, the half-life of a sample of a gaseous compound initially at 56.0 kPakPa is 340s. When the pressure is 28.0 kPakPa, the half-life is 170s. The order of the reaction is:
    Solution
    When the pressure is reduced to half from 56.0 kPa to 28 kPa, the half-life period is reduced to half from 340 s to 170 s. 

    Thus, the half-life period is proportional to the pressure. This is true for a zero-order reaction.
  • Question 4
    1 / -0
    N2O2(g)2NON_{2}O_{2} (g)\rightarrow2NO   is  a first-order reaction interms of the concentration of N2O2(g)N_{2}O_{2} (g)  .Which of the following is valid, [N2O2][N_{2}O_{2} ]  being constant?
    Solution
    The reaction is N2O2(g)2NON_{2}O_{2} (g)\rightarrow2NO
    The initial concentration of  N2O2N_{2}O_{2} is [N2O2]0[N_{2}O_{2}]_0.. Assuming initially only N2O2N_{2}O_{2} is present and no NONO is formed.
    At time t, the concentrtion of N2O2N_{2}O_{2} and NO are [N2O2][N_{2}O_{2}] and [NO][NO] respectively.
    The total concentration is equal to [N2O2]+[NO]=[N2O2]0[N_{2}O_{2}]+[NO]=[N_{2}O_{2}]_0  ......(1)
    But from integrated rate law equation fo first order reaction , [N2O2]=[N2O2]0ekt[N_{2}O_{2}]=[N_{2}O_{2}]_0e^{-kt}......(2)
    Substituting equation (2) in equation (1)
    [N2O2]0ekt+[NO]=[N2O2]0[N_{2}O_{2}]_0e^{-kt}+[NO]=[N_{2}O_{2}]_0
    [NO]=[N2O2]0[N2O2]0ekt=[N2O2]0(1ekt)[NO]=[N_{2}O_{2}]_0-[N_{2}O_{2}]_0e^{-kt}=[N_{2}O_{2}]_0(1-e^{-kt})
  • Question 5
    1 / -0
    The time required for the completion of first order reaction is:
    Solution
    k=1t×ln(aax)k= \cfrac{1}{t}\times ln\left(\cfrac{a}{a-x}\right)

    For a First-order reaction, we have

    (ax)(a-x) =aekt= ae^{-kt}

    Where (ax)(a-x) is the concentration of reactant AA at any time, tt
     
    When the reaction goes to completion 

    (ax)=0(a-x)=0 and we get,

    0=aekt0= ae^{-kt}

    t=t=\infty

    Hence, the correct option is A\text{A}
  • Question 6
    1 / -0
    The reaction LML\rightarrow M is started with 10.0 gm10.0\ gm of LL. After 3030 and 9090 minutes, 5.0 gm5.0\ gm and 1.25 gm1.25\ gm of LL respectively are left. The order of the reaction is:
    Solution
    After 30 minutes, the concentration is reduced to half.

    After the next 60 minutes, the concentration is reduced to one fourth.

    Thus, 30 minutes is the half-life period and is independent of the concentration of LL.

    Hence, it is a first-order reaction.

    Option B is correct.
  • Question 7
    1 / -0
    For the reaction A \rightarrow B it has been found that the order of the reaction is zero with respect to A. Which of the following expressions correctly describes the reaction?
    Solution
    For a zero order reaction,  d[A]=Kdt.d[A]=-Kdt.

    On integrating this equation between limits, 

    [A]=[A]0[A]=[A]_0 at t=0 t=0    

    [A]=[A]t [A]=[A]_t  at t=t t=t, we get [A0][A]=Kt[A_{0}] - [A] = Kt
  • Question 8
    1 / -0
    SO2Cl2 SO2+Cl2SO_{2}Cl_{2} \rightarrow  SO_{2}+Cl_{2} is a first order gas reaction with k=2.2×105sec1k=2.2\times 10^{-5}\sec^{-1} at 320oC320^o C. The percentage of SO2Cl2SO_{2}Cl_{2} decomposed on heating for 90 minutes is:
    Solution
    For a first order reaction,

    k=ln[RoR]tk=\dfrac{ln\left[\dfrac{Ro}{R} \right ]}{t}

    ln[RoR]=2.2×105×90×60ln\left[\dfrac{Ro}{R} \right ]=2.2\times10^{-5}\times90\times60

    lnRoR=0.1188ln\dfrac{Ro}{R}=0.1188

    RoR=1.126\dfrac{Ro}{R}=1.126

    R=0.8879RoR = 0.88 79 Ro

    Percentage of decomposition =  R0R Ro0.8879 RoR_0-R\ \Rightarrow Ro-0.8879\ Ro
    11.18\Rightarrow 11.18 RoRo

    Hence, option D is correct.
  • Question 9
    1 / -0
    Which equation represents the time to complete 90% of first order reaction?
    Solution
    t=2.303k log[RoR]t = \cfrac{2.303}{k}\ log \left [ \cfrac{Ro}{R} \right ]

    t=2.303k log [aa0.9a ]t =\cfrac{2.303}{k}\ log\ \left [ \cfrac{a}{a-0.9a} \right ]

    t=2.303k log[10.1]t = \cfrac{2.303}{k}\ log \left [ \cfrac{1}{0.1} \right ]

    =2.303k= \cfrac{2.303}{k}

    Hence, the correct option is C\text{C}
  • Question 10
    1 / -0
    At room temperature the reaction between NO and O2O_{2} to give NO2NO_{2} is fast while that of between CO and O2O_{2} is slow. It is because:
    Solution
    The reaction between NO and oxygen has lower activation energy than the reaction between CO and oxygen.
    Hence, the reaction between NO and oxygen is fast whereas the reaction between CO and oxygen is slow.
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