Trace methane in the Earth’s lower atmosphere contributes to the greenhouse effe
ID: 497702 • Letter: T
Question
Trace methane in the Earth’s lower atmosphere contributes to the greenhouse effect. Methane absorbs infrared radiation from the Earth’s surface leading to potential global warning. The main pathway by which methane is removed from the troposphere is by reaction with the hydroxyl radical:
4. Trace methane in the Earth's lower atmosphere contributes to the greenhouse effect. Methane absorbs infrared radiation from the Earth's surface leading to potential global warning. The main pathway by which methane is removed from the troposphere is by reaction with the hydroxyl radical CH4 H2OB) CH3(g) (g) The activation energy for this reaction is 19.5 kJ/mole. The temperature of the troposphere declines steadily with altitude to about 10 km above sea-level. At some intermediate altitude the temperature is -14°C. What is the ratio of the rate constant for the reaction at this temperature over that for the reaction at 250C?Explanation / Answer
Here we use Arrhenius equation: k=Ae-Ea/RT
Where, k=rate constant;
A=frequency factor or pre-exponential factor
Ea=Activation energy
R=universal gas constant=8.314 JK-1mol-1
T=temperature
Now, for two different rate constants k1 and k2 we have:
K1=Ae-Ea/RT1….(i)
K2=Ae-Ea/RT2…..(ii)
We get the ratio of k1 and k2 by dividing equation (i) by (ii)
K1/K2 = [Ae-Ea/RT1]/ [Ae-Ea/RT2]
K1/K2 = e-Ea/RT1/ e-Ea/RT2 (*A factor gets cancelled out)
K1/K2 = e-Ea (1/T1-1/T2)/R
K1/K2 = eEa (1/T2-1/T1)/R….(iii)
We are given T1= -14 oC= [273+(-14)] K = 259 K
T2= 25 oC= [273+25] K = 298 K
Ea= 19.5 kJmol-1=19500 Jmol-1
Putting the values of T1, T2, R and Ea in equation (iii) we get:
K1/K2 = e19500 (1/298-1/259)/8.314
K1/K2 = e-1.185
K1/K2 = 0.3057
Therefore, the ratio of the rate constant at -14 oC over that at the reaction at 25 oC is 0.3057.
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