Dear Student, The titre value at zero time may be taken as initial concentration i e a=22 8At different intervals, the value of a-x are givenk=2 303tlogaa-x(i) k1=2 303600log22 813 8=8 36×10-4(ii) k2=2 3031200log22 88 20=8 52×10-4Overall, k=k1+k22=8 36×10-4+8 52×10-42=8 44×10-4 s

Answer is A

Answer is A Q. 19 decomposition of hydrogen peroxide in an aqueous solution is a reaction of first order. It can be followed by titrating 10 ml portions of reaction mixture at various times from the beginning of reaction against a standard solution of KMn04. Volume of KMn04 solution used in each case is proportional to the remaining concentration of From the following data, calculate the rate constant of the reaction. Time (in second) 0 600 1200 KMnoasol. used (in ml) 22.8 13.8 8.20 (A) 8.44 x sec-I (B) k — 8.44 x 10-2 sec-I (C) k- 2.44 sec-I (D) k -2.22 x 104 sec-I Effect of temperature on rate based on constant (Arrhenius equation) Q.20 distribution Of molecular kinetic energy at teniperature is as shown in the following graph. ,

Dear Student,

The titre value at zero time may be taken as initial concentration i . e . a = 22.8 At different intervals, the value of a - x are given k = \frac{2.303}{t} \log \frac{a}{a - x} (i) k_{1} = \frac{2.303}{600} \log \frac{22.8}{13.8} = 8.36 \times 10^{- 4} (ii) k_{2} = \frac{2.303}{1200} \log \frac{22.8}{8.20} = 8.52 \times 10^{- 4} Overall, k = \frac{k_{1} + k_{2}}{2} = \frac{8.36 \times 10^{- 4} + 8.52 \times 10^{- 4}}{2} = 8.44 \times 10^{- 4} s

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