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CHEMISTRY6. From the following data, form the reaction betweenA and B.[IIT-1994][A] [B] Initial rate (mol L –1 s –1 )mol L –1 mol L –1 300 K 320 K2.5 ×10 –4 3.0 ×10 –5 5.0 ×10 –4 2.0 × 10 –35.0 × 10 –4 6.0 × 10 –4 4.0 × 10 –3 –1.0 × 10 –3 6.0 × 10 –5 1.6 × 10 –2 –Calculate(a) the order of reaction with respect to A and withrespect to B,(b) the rate constant at 300 K,(c) the energy of activation,(d) the pre exponential factor.Sol. Rate of reaction = k[A] l [B] mwhere l and m are the order of reaction with respectto A and B respectively. From the given data, weobtain following expressions :5.0 × 10 –4 = k[2.5 × 10 –4 ] l [3.0 × 10 –5 ] m ...(i)4.0 × 10 –3 = k[5.0 × 10 –4 ] l [6.0 × 10 –5 ] m ...(ii)1.6 × 10 –2 = k[1.0 × 10 –3 ] l [6.0 × 10 –5 ] m ..(iii)From eq. (ii) and eq. (iii), we get4.0×101.6×10−3−2⎛ 5.0×10= ⎜⎝ 1.0×10−4−3l⎞⎟⎠or 0.25 = (0.5) lor (0.5) 2 = (0.5) lor l = 2From eq. (i) and eq. (ii), we getor5.0×101.0×10−4−3⎛ 2.5 × 10= ⎜⎝ 5.0 × 10m1 1 ⎛ 1 ⎞= × ⎜ ⎟⎠8 4 ⎝ 2−4−42⎞ ⎛ 3.0 × 10⎟ ⎜⎠ ⎝ 6.0 × 10−5−5m1 ⎛ 1 ⎞or = ⎜ ⎟⎠2 ⎝ 2or m = 1(b) At T 1 = 300 K,−4Rate of reaction 5.0×10k 1 = =2 1−42[A] [B] [2.5×10 ] [3.0×10= 2.67 × 10 8 L 2 mol –2 s –1(c) At T 2 = 320 K,Rate of reactionk 2 =2 1[A] [B]−32.0×10=−42−5[2.5×10 ] [3.0×10 ]= 1.067 × 10 9 L 2 mol –2 s –1We know, 2.303 logkk21=E ⎛ ⎞a T2− T1⎜⎟R ⎝ T1T2 ⎠⎞⎟⎠m−5]91.067 × 10 ⎛ 320 − 300 ⎞or 2.303 log =8⎜ ⎟2.67 × 10 8.314 ⎝ 320×300 ⎠E aEor 2.303 × 0.6017 = a ⎛ 20 ⎞⎜ ⎟8.314 ⎝ 320×300 ⎠2 .303×0.6017×8.314×320×300or E a =20= 55.3 kJ mol –1(d) According to Arrhenius equation,/ RTk = Ae −E aor 2.303 log k = 2.303 log A – RTAt 300 K,2.303 log (2.67 × 10 8 355.3×10) = 2.303 log A –8.314×300or 2.303 × 8.4265 = 2.303 log A – 22.1719 .4062 + 22.17 41.5762or logA == = 18.05312.303 2.303A = Antilog 18.0531 = 1.13 × 10 18 s –17. 1 g of a mixture containing equal number of moles ofcarbonates of two alkali metals, required 44.4 ml of0.5 N HCl for complete reaction. The atomic weightof one metal is 7, find the atomic weight of othermetal. Also calculate the amount of sulphate formedon quantitative conversion of 1.0 g of the mixture intwo sulphates.[IIT-1972]Sol. Let, Mass of one alkali metal carbonate M 2 CO 3 = xgThen, mass of other alkali metal carbonate M 2´CO 3= (1 – x)gStep 1.Molecular mass 74Equivalent mass of M 2 CO 3 ==2 2= 37 (at mass of M = 7)E aMeq. of M 2 CO 3 = 37x ×1000xMoles of M 2 CO 3 = 74Step 2Molecular massEquivalent mass of M 2´CO 3 =22 m + 60=2( m = atomic mass of M´)2(1 − x)Meq. of M 2´CO 3 = × 10002m + 601−xMoles of M 2´CO 3 =2m + 60Step 3. Meq. of HCl = N HCl × V HCl = 0.5 × 44.4Step 4. According to the question,Moles of M 2 CO 3 = Moles of M 2´CO 3x 1−xor = ...(i)74 2m + 60XtraEdge for IIT-JEE 10 APRIL 2010