Question #fa922

1 Answer
Jun 29, 2017

Answer:

Here's what I get.

Explanation:

(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#