TYPE OR PRINT!!!!!!! v. The following set of data was obtained by the method of
ID: 705392 • Letter: T
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TYPE OR PRINT!!!!!!!
v. The following set of data was obtained by the method of initial rates for the reaction: 2 HgCl2 (aq) + C,042. (aq) 2 Cli. (aq) + 2 CO2 (g) + H&CI2; (s) What is the rate law for the reaction? [HgCl2], M 0.10 0.10 0.20 Rate 0.10 0.20 0.20 Rate, M/s 1.3 x 10-7 5.2 x 10-7 1.0 x 10-6 k[reactant A]Treactant B]" so the Rate law would be Rate -k[HgCl, c,02 ??? not sure how to get n and m variables I had to look this answer up vi. The reaction for the decomposition of dinitrogen monoxide gas to form oxygen radicals is: N20 (g) > N2 (g) + 0 (g). If the rate constant, k, is 3.04 x 10-2 s-1 and the frequency factor, A, is 8.00 x 1011 s1, what is the activation energy for the first-order reaction at 700 °C?Explanation / Answer
Let Rate r= K[HgCl2] m[ C2O4-2] n, where m and n are orders of reaction with respect to HgCl2 and C2O4-2] respectively.K is the rate constant.
From the 1st data point, K[0.1]m [0.1]n= 1.3*10-7, (1)
RFrom the 2nd data point, K[0.1]m[0.2]n= 5.2*10-7 (2) and
Eq.2/Eq.1 gives K[0.1]m[0.2]n/ K(0.1)m *(0.1)n= 5.2/1.3= 4, 2n=4 and n=2
From 3rd data point, K[0.2]m[0.2]n= 1*10-6 (3)
Eq.3/Eq.2 gives 2m= 1*10-6/(5.2*10-7)= 1.92, m=0.94 ( order since is determined experimentally can be fraction as well)
So the rate equation becomes , r= K[HgCl2]0.94[C2O4-2]2
From Eq.1, K[0.1]0.94[0.1]2= 1.3*10-7, K=0.000113/sec.M1.94
2. Arhenius equation is K= Ao*e(-E/RT) or lnK= lnKo-E/RT (1)
K is rate constant, Ao is frequency factor, E is activation energy , R= gas constant= 8.314 J/mole.K and T is temperature in K. Given K= 3.04*10-2/sec, Ao= 8*1011/sec, T= 700 deg.c= 700+273= 973K
substituting these values in Eq.1 gives ln(3.04*10-2)= ln(8*1011)- E/(8.314*973)
E= 249976 J/mole=249.976 Kj/mole
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