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Rate law of the n^(th) order reaction is...

Rate law of the `n^(th)` order reaction is given by `(dx)/(dt)=k[A]^(n)`
What kinetic parameters can you derive from logarithm graph

A

n(order) of reaction

B

k(rate constant)

C

Both a and b

D

None of these

Text Solution

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The correct Answer is:
To derive kinetic parameters from the logarithmic graph of the rate law for an nth order reaction given by \((dx)/(dt) = k[A]^n\), we can follow these steps: ### Step 1: Write the Rate Law The rate law for the nth order reaction is given by: \[ \frac{dx}{dt} = k[A]^n \] where: - \(\frac{dx}{dt}\) is the rate of the reaction, - \(k\) is the rate constant, - \([A]\) is the concentration of reactant A, - \(n\) is the order of the reaction. ### Step 2: Take the Logarithm To analyze this equation using a logarithmic graph, we can take the logarithm of both sides: \[ \log\left(\frac{dx}{dt}\right) = \log(k) + n \log([A]) \] ### Step 3: Rearrange the Equation This equation can be rearranged into the form of a straight line \(y = mx + c\): - Let \(y = \log\left(\frac{dx}{dt}\right)\) - Let \(x = \log([A])\) - The slope \(m = n\) - The y-intercept \(c = \log(k)\) Thus, we have: \[ \log\left(\frac{dx}{dt}\right) = n \log([A]) + \log(k) \] ### Step 4: Analyze the Graph From the graph of \(\log\left(\frac{dx}{dt}\right)\) versus \(\log([A])\): - The slope of the line gives the order of the reaction \(n\). - The y-intercept gives the logarithm of the rate constant \(\log(k)\). ### Conclusion From the logarithmic graph, we can derive: 1. The order of the reaction \(n\). 2. The rate constant \(k\).

To derive kinetic parameters from the logarithmic graph of the rate law for an nth order reaction given by \((dx)/(dt) = k[A]^n\), we can follow these steps: ### Step 1: Write the Rate Law The rate law for the nth order reaction is given by: \[ \frac{dx}{dt} = k[A]^n \] where: ...
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The rate of the reaction A to Products is given by (dx)/(dt)=k[A]^(1//2) . The units of the rate constants are:

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The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of nth order reaction is k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))] …(i) Half life of nth order reaction depends on the initial concentration according to the following relation: t_(1//2) prop (1)/(a^(n-1)) ...(ii) The unit of the rate constant varies with the order but general relation for the unit of nth order reaction is Units of k = [(1)/(Conc)]^(n-1) xx "Time"^(-1) ...(iii) The differential rate law for nth order reaction may be given as: (dx)/(dt) = k[A]^(n) ...(iv) where A denotes the reactant. In a chemical reaction A rarr B , it is found that the rate of the reaction doubles when the concentration of A is increased four times. The order of the reaction with respect to A is:

The rate law of the reaction A+2Brarr Product is given by (d("Product"))/(dt)= k[A]^(2)[B] . A is taken in large excess, the order of the reqaction will be

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Knowledge Check

  • The rate of the reaction A to Products is given by (dx)/(dt)=k[A]^(1//2) . The units of the rate constants are:

    A
    `lit mol^-1 sec^-1`
    B
    `lit^(1//2) mol^(-1//2) sec^-1`
    C
    `lit^(-1//2) mol^(1//2) sec^-1`
    D
    `lit^(3//2) mol^(-1//2) sec^-1`
  • The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of nth order reaction is k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))] …(i) Half life of nth order reaction depends on the initial concentration according to the following relation: t_(1//2) prop (1)/(a^(n-1)) ...(ii) The unit of the rate constant varies with the order but general relation for the unit of nth order reaction is Units of k = [(1)/(Conc)]^(n-1) xx "Time"^(-1) ...(iii) The differential rate law for nth order reaction may be given as: (dX)/(dt) = k[A]^(n) ...(iv) where A denotes the reactant. The rate constant for zero order reaction is where c_(0) and c_(t) are concentration of reactants at respective times.

    A
    `k = (c_(0))/(2t)`
    B
    `k = (c_(0)-c_(t))/(t)`
    C
    `k = ln.(c_(0)-c_(t))/(2t)`
    D
    `k = (c_(0))/(c_(t))`
  • The rate of law of the reaction A+2B to Product is given by (d("product"))/(dt)=k[A]^(2).[B] If A is taken in large excess , the order of the reaction will be

    A
    0
    B
    1
    C
    2
    D
    3