Consider the reaction A_2(g) + X_2(g) rightarrow products. Two trials were perfo

Question

Consider the reaction A_2(g) + X_2(g) rightarrow products. Two trials were performed at the same temperature and in the same container. The initial rate was found to increase by a factor of 2.83 when the molar concentration of X_2 was halved and the molar concentration of A_2 was unchanged. (a) Can you determine the order with respect to X_2 using the information given above? If yes, determine the order. If no, explain why and state what additional information would be needed to determine the order with respect to X_2. (b) Can you determine the order with respect to A_2 using the information given above? If yes, determine the order. If no, explain why and state what additional information would be needed to determine the order with respect to A_2.

Explanation / Answer

a)

A2+X2------> Product

Rate equation will be , ra=k*Caa*Cxb               ---------------------------------------(1)

Given,

Process conditions are constant

The initial rate was increased by factor 2.83 when molar concentration of X2 is halved keeping [Ca] constant.

Therefore,

(ra1/ra2)= (1/2.83)=0.3533

(Ca1/Ca2)=1

(Cx1/Cx2)= (1/2)=0.5

Note: Subscript 1 & 2 indicate experiment no.

From equation (1),

For experiment 1 :     ra1=k*Ca1a*Cx1b

For experiment 2 :     ra2=k*Ca2a*Cx2b

Dividing ,

(ra1/ra2)= (k*Ca1a*Cx1b)/( ra2=k*Ca2a*Cx2b)

0.3533= (k/k)* (Ca1/Ca2)a (Cx1/Cx2)b

0.3533 = 1*(1)*(0.5)b

0.3533=(0.5)b

Log 0.3533=b*log 0.5

b=1.5

b)

With given data it is not possible to calculate order of reaction with respect to A2.

Addition data required : initial rate change factor with changing molar concentration of A2 & keeping molar concentration X2 Constant or changing.

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