Consider the dissociation of methane, CH_4 (g), into the elements H_2 (g) and C(
ID: 1048774 • Letter: C
Question
Consider the dissociation of methane, CH_4 (g), into the elements H_2 (g) and C(s, graphite). Given that delta_f H degree (CH_4, g)=-74.85 kJ/mol and that delta_f S degree (CH_4, g)=-80.67 J/Kmol at 298 K, calculate the value of the equilibrium constant at 298 K. Assuming that delta_f H degree is independent of temperature, calculate K at 50 degree C. Calculate the degree of dissociation, alpha, of methane at 25degree C and total pressure of 0.01 bar. Without doing any numerical calculations, explain how the degree of dissociation for this reaction will change as the pressure and temperature are varied.Explanation / Answer
we know that detlaG= deltaH- TdeltaS
deltaG= 74.85*1000 - 298*80.67=50810.34 j/mol= 50.810 Kj/mol
but deltaG= -RT lnK
50810.34= -8.314*298*ln K
lnK= -20.51
K= 1.24*10-9
2. at 50 deg.c= 50+273.15= 323.15K
ln (K2K1)= (deltaH/R)*(1/T1-1/T2)
K2= equilibrium constant at T2= 50 deg.c= 323.15 K, K1= Equilibrium constant at T1= 298K = 1.24*10-9
ln (K2/K1)= (74.85*1000/8.314)*(1/298-1/323.15)
K2= 1.30*10-8
CH4 (g)----> C + 2H2(g)
at 0.01 bar and 298K
from PV= nRT
n/V= concentration = P/RT= 0.01*0.9865 atm/(0.0821*298)=0.000403M
let x = drop in concentratin of CH4 at equilibrium
So at equilibrium , CH4= 0.000403-x [H2] =2x
Kc= (2x)2/ (0.00403-x)= 1.24*10-9
x2/(0.00403-x)= (1.24*10-9)/4 =3.1*10-10
when solved using excel, x= 1.12*10-6 M
% dissociation= 100*1.12*10-6/ 0.00403 =0.028%
d) CH4 -------->C+2H2
the reaction is endothermic
So when temperature is increased, according to Lechatlier principle, the effect of increase in temmperature is compensated in the forward direction. So degree of dissociaiton increases with an increase in temperature.So increase in dissociation.
when pressure is increased, the reaction proceeds in a direction where there is decrease in number of moles so as to compensate the effect of increase in pressure. So backward direction is favored. So decrease in dissociation.