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The following data were obtained during the first order thermal decomposition of SO2 Cl2 at constant volume.
   SO2 Cl2 (g) → SO2 (g) + Cl2 (g)

Experiment

Time/s-1

Total pressure/atm

1

0

0.5

2

100

0.6

Calculate the rate of reaction when total pressure is 0.65 atm.

we have give the data

K = 2.303tlogp1p2     = 2.303100×log0.50.6or K = 2.23 × 10-35-1

Rate of reaction = 7.8 × 10-4 atm s-1
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The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K, if the value of A is 4 x 1010 s–1. Calculate k at 318 k and Ea


For the first order reaction 

(i) t10% = 2.303klog10090         = 2.303klog1009

(ii) t25% = 2.303klog10075       = 2.303k1log43

According to problem, t10% = t25%
Therefore,
               2.303k1log109 = 2.303k2log431k1×log 10 × log 9 = 1k2×0.6021-0.4771   1k1×1-0.9542 = 1k2×0.602-0.4771       k1k2= 0.6021-0.47711-0.9542               = 0.1250.0458 = 2.7
Form the Arrhenius equation, we obtain

logk1k2 = Ea2.303 R1T1-1T2log 2.7 = Ea2.304×8.314 R1298-13080.4314 = Ea2.304×8.314 R×(308-298)298×308or 0.4314 = Ea2.304×8.314 R×10298×308Ea = 0.4314×2.304×8.314×298×30810      = 75.847 kJ mol-1.


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Consider a certain reaction A → Products with K = 2.0 x 10–2 s–1. Calculate the concentration of A remaining after 100 S if the initial concentration of A is 1.0 mol L–1.

k= 2.0 x10-2 s-1
T=100
[A]0 =1.0 mol-1
The givaen reaction is first order reaction

K = 2.0 × 10-2s-1t1/2 = 100 sec, a = 1.0 mol L-1K = 2.303tlogaa-x2.0×10-2 = 2.303100log1.01.0-xor 2.0×10-2×1002.303=log11-xor  0.8684 = log11-xAntilog 0.8684 = log11-xor 7.386 = 11-xor 7.386(1-x) = 1or 1-x=17.386=0.135

Hence, concentration of A remaining after 100 S = 0.135 M.
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The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponental factor. 

According to the Arrhenius equation 

K = 2.418 × 10-5 S-1Ea = 179.9 kJ/mol.K = Ae-Ea/RTlog K = log A - Ea2.303RlogA = log 0.0000241+179.92.303×8.314×546logA = log 0.0000241+0.02A = 3.9 × 162s-1.
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Sucrose decomposes in acid solution into glucose and fuctose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

We have given that

t1/2 = 3.00 hours.For the first order reaction,t1/2 = 0.693Kor K = 0.693t1/2=0.6933=0.231 hr-1Now K = 2.303tlogaa-x0.231 = 2.3038logaa-x0.231×82.303 = logaa-x0.8024 = log aa-xor aa-x=Antilog 0.8024aa-x=6.345or fraction left = a-xa=16.345=0.157 M.
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