(a) #M_text(R)# as a function of time
This is really a disguised first-order kinetics problem.
#S# is the rate of gas supply, and it depends on the mass #M# of gas.
The equation for this first order reaction is
#"Rate" = S = "-"(dM)/(dt) = AM#
where
#A# is the rate constant and
#M# is the mass of gas available
We can rewrite this as
#(dM)/M = "-"Adt#
If we integrate this expression, we get
#int(dM)/M = lnM = "-"At + C#, where #C# is a constant of integration.
At #t = 0, M = M_i#, so #C= lnM_i#
At any time #t#, #M# is the mass remaining, #M_R#.
∴ The integrated rate law in logarithmic form is
#lnM_R = lnM_i - At#.
In exponential form, we can write this as
#color(blue)(bar(ul(|color(white)(a/a)M_R = M_ie^("-"At)color(white)(a/a)|)))" "#
(b) Percentage of gas that came from this field
#M_R = M_i × e^("-"0.2color(white)(l) "yr"^"-1" × 10color(white)(l) "yr") = M_i × e^"-2" = 0.14 M_i#
So, 14 % of the original mass remains after 10 yr.
∴ 86 % of the original mass was consumed (the rest must have come from some other field).
(c) Thermal energy released
#"Total energy" = mΔ_cH = 2.6 × 10^6 color(red)(cancel(color(black)("t"))) × (1000 color(red)(cancel(color(black)("kg"))))/(1 color(red)(cancel(color(black)("t")))) × "56 MJ"/(1 color(red)(cancel(color(black)("kg")))) = 1.5 × 10^11color(white)(l) "MJ"#
(d) Mass of #"CO"_2#
#M_text(R): color(white)(m) 16 color(white)(mmmmmml)44#
#color(white)(mmm)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"#
#"Mass of CO"_2 = 2.6 × 10^6 color(red)(cancel(color(black)("t CH"_4))) × (44 color(white)(l)"t CO"_2)/(16 color(red)(cancel(color(black)("t CH"_4)))) = 7.2 × 10^6 color(white)(l)"t CO"_2#
