Home
Class 12
CHEMISTRY
Carbon monoxide reacts with O(2) to form...

Carbon monoxide reacts with `O_(2)` to form `CO_(2)`: `2CO(g)+O_(2)(g) to 2CO_(2)(g)`
Infromations about this reaction are given in the table below.
`{:([CO]"mol/L",[O_(2)]"mol/L", "Rate of reaction (mol/L.min)"),(0.02,0.02,4xx10^(-5)),(0.04,0.02,1.6xx10^(-4)),(0.02,0.04,8xx10^(-5)):}`
What is the value for the rate constant for the reaction in properly related unit?

Text Solution

AI Generated Solution

The correct Answer is:
To find the rate constant (k) for the reaction of carbon monoxide with oxygen to form carbon dioxide, we will follow these steps: ### Step 1: Determine the Rate Law The general form of the rate law for the reaction \(2CO(g) + O_2(g) \rightarrow 2CO_2(g)\) can be expressed as: \[ \text{Rate} = k [CO]^m [O_2]^n \] where \(m\) and \(n\) are the orders of the reaction with respect to \(CO\) and \(O_2\), respectively. ### Step 2: Analyze the Data From the provided data: 1. For \([CO] = 0.02 \, \text{mol/L}, [O_2] = 0.02 \, \text{mol/L}, \text{Rate} = 4 \times 10^{-5} \, \text{mol/L.min}\) 2. For \([CO] = 0.04 \, \text{mol/L}, [O_2] = 0.02 \, \text{mol/L}, \text{Rate} = 1.6 \times 10^{-4} \, \text{mol/L.min}\) 3. For \([CO] = 0.02 \, \text{mol/L}, [O_2] = 0.04 \, \text{mol/L}, \text{Rate} = 8 \times 10^{-5} \, \text{mol/L.min}\) ### Step 3: Determine the Order with Respect to \(O_2\) Compare the first and third experiments where \([CO]\) is constant: - From \(0.02 \, \text{mol/L}\) to \(0.04 \, \text{mol/L}\) (doubling \(O_2\)): \[ \text{Rate} \text{ changes from } 4 \times 10^{-5} \text{ to } 8 \times 10^{-5} \text{ (doubles)} \] This indicates that the reaction is first order with respect to \(O_2\) (n = 1). ### Step 4: Determine the Order with Respect to \(CO\) Now, compare the first and second experiments where \([O_2]\) is constant: - From \(0.02 \, \text{mol/L}\) to \(0.04 \, \text{mol/L}\) (doubling \(CO\)): \[ \text{Rate} \text{ changes from } 4 \times 10^{-5} \text{ to } 1.6 \times 10^{-4} \text{ (quadruples)} \] This indicates that the reaction is second order with respect to \(CO\) (m = 2). ### Step 5: Write the Rate Law The overall rate law can now be expressed as: \[ \text{Rate} = k [CO]^2 [O_2]^1 \] ### Step 6: Calculate the Rate Constant \(k\) Using the first set of data: \[ 4 \times 10^{-5} = k (0.02)^2 (0.02) \] Calculating: \[ 4 \times 10^{-5} = k (0.0004)(0.02) \] \[ 4 \times 10^{-5} = k (8 \times 10^{-6}) \] Now, solving for \(k\): \[ k = \frac{4 \times 10^{-5}}{8 \times 10^{-6}} = 5 \, \text{L}^2/\text{mol}^2 \cdot \text{min} \] ### Final Answer The value of the rate constant \(k\) is: \[ k = 5 \, \text{L}^2/\text{mol}^2 \cdot \text{min} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

The reaction of A_(2) and B_(2) follows the equation A_(2)(g)+B_(2)(g)to2AB(g) The following data were observed {:(,[A_(2)]_(0),[B_(2)]_(0),"Initial rate of appearance of AB(g)(in" Ms^(-1)),(,0.10,0.10,2.5xx10^(-4)),(,0.20,0.10,5xx10^(-4)),(,0.20,0.20,10xx10^(-4)):} The value of rate constatnt for the above reaction is :

Calculate the order of the reaction in A and B :- {:(A,,,B,,,Rate),(("mol"//1),,,("mol"//1),,,),(0.05,,,0.05,,,1.2xx10^(-3)),(0.10,,,0.05,,,2.4xx10^(-3)),(0.05,,,0.10,,,1.2xx10^(-3)):}

In data given below. {:(,[A],[B],"rate"),(,mol//L,mol//L,mol L ^(-1)s^(-1)),(a., 0.02,1.2,3.0xx10^(-3)),(b.,0.04,2.4,6.0xx10^(-3)):} The reaction may be

For a hypothetical reaction A + B rarr C , suggest the rate law and order form the following data: |{:("Experiment",[A] (mol L^(-1)),[B] (mol L^(-1)),"Refer of reaction" (mol L^(-1) s^(-1)),),(I,0.25,0.25,3.0 xx 10^(-3),),(II,0.50,0.25,6.0 xx 10^(-3),),(III,0.50,0.50,1.20 xx 10^(-2),):}|

For a reaction 2NO(g) + 2H_(2)(g) rarr N_(2)(g)+2H_(2)O(g) the following data were obtained: |{:(,[NO] (mol L^(-1)),[H_(2)] (mol L^(-1)),"Rate" (mol L^(-1) s^(-1))),(1.,5 xx 10^(-3),2.5 xx 10^(-3),3 xx 10^(-5)),(2.,15 xx 10^(-3),2.5 xx 10^(-3),9xx10^(-5)),(3.,15 xx 10^(-3),10 xx 10^(-3),3.6 xx 10^(-4)):}| (a) Calculating the order of reactions. (b) Find the rate constant. (c ) Find the initial rate if [NO] = [H_(2)] = 8.0 xx 10^(-3) M

The kinetic date for the given reaction A(g)+2B(g)overset(K)toC(g) is provided in the following table for three experiments at 300 K {:("Ex. No.","[A/M]","[B/M]","Initial rate" (M sec^(-1))),(1.,0.01,0.01,6.930xx10^(-6)),(2.,0.02,0.01,1.386xx10^(-5)),(3.,0.2,0.02,1.386xx10^(-5)):} In another experiment starting swith initial concentration of 0.5 and 1 M respectively for A and B at 300 K, find the rate of reaction after 50 minutes from the starting of edperiment (inM // sec).

Consider the reaction, 2N_(2) O_(5) to 4NO_(2) + O_(2) In the reaction NO_(2) is being formed at the rate of 0.0125 mol L^(-1) s^(-1) . What is the rate of reaction at this time?

Rate of a reaction A + B rarr Product, is given as a function of different initial concentration of A and B . |{:([A] (mol L^(-1)),(B) (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}| Determine the order of the reaction with respect to A and with respect to B . What is the half life of A in the reaction ?

The rate constant of a first order reaction is 4.5 xx 10^(-2) sec^(-1) What will be the time required for the initial concentration of 0.4 M of the reactant to be reduced to 0.2 M ?