For the reaction X + Y ? Products, the following data were obtained. (HINT: Reac

Question

For the reaction X + Y ? Products, the following data were obtained. (HINT: Reaction times are inversely proportional to reaction rates.) exp. no Y reaction time (s) 0.220 0.440 0.439 0.440 0.100 0.102 0.050 0.100 0.224 0.224 0.223 0.447 214 107 108 54 the rate law, including reaction orders with respect to all reactants, and show how you obtained the reaction orders. change in some way to get baca t ducing more B and C and using up some A. It must do us, values that, on substitution into Equation 2, equal 10. At that point the system us had been no forced shift to the right. an t. An increase in concentration of a product will ntration causes a shift ljbrium state, [B] and (C) are greater than they were initially, and [A] is larger than is that one can always cause a reaction to ilibrium n reading the last

Explanation / Answer

Sol. As reaction rate is inversely proportional to the reaction time , so , reaction rate can be written as the reciprocal of reaction time .

Let the orders of reaction with respect to X,Y,Z are m ,n , o respectively ,so ,

Reaction Rate = K[X]m[Y]n[Z]o (equation 1)

  where K = rate constant .

Now , taking all the data of experiment no. 1 and apply to equation no. 1 ,

1/214 = K (0.220)m (0.100)n (0.224)o (equation 2)

Taking all the data of experiment no. 2 and apply to equation no. 1 ,

1/107 = K(0.440)m(0.102)n(0.224)o (equation 3 )

Taking all the data of experiment no. 3 and apply to equation no. 1 ,

1/108 = K(0.439)m(0.050)n(0.223)o (equation 4)

Taking all the data of experiment no. 4 and apply to equation no. 1 ,

1/54 = K(0.440)m(0.100)n(0.447)o (equation 5)

Now , dividing equation 3 by 2 , we have ,

214/107 = (0.440/0.220)m(0.102/0.100)n(0.224/0.224)o ,as the 0.102 is approximately equal to 0.100 , so ,

2 = (2)m(1)n(1)o = (2)m which implies m = 1.

Similarly , dividing eq. 4 by 3 , we have ,

107/108 = (0.05/0.102)n(1)m(1)o= (0.5)n (as 0.223 is approximate equal to 0.224 and 0.439 is approximate equal to 0.440)

1 = (0.5)n => (0.5)0 = (0.5)n which implies n = 0.

Similarly , dividing equation 5 by 3 , we have ,

108/54 = (0.447/0.224)o (1)m(1)n ( as 0.102 is approximate equal to 0.100 )

2 = (2)o which implies o = 1.

Therefore , Reaction Rate = K[X]1[Y]0[Z]1 and the oder of reaction with respect to X ,Y ,Z are 1 ,0 and 1 respectively .

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