# Question fa922

Jun 29, 2017

Here's what I get.

#### Explanation:

(a) ${M}_{\textrm{R}}$ 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

$\text{Rate" = S = "-} \frac{\mathrm{dM}}{\mathrm{dt}} = A M$

where

$A$ is the rate constant and
$M$ is the mass of gas available

We can rewrite this as

$\frac{\mathrm{dM}}{M} = \text{-} A \mathrm{dt}$

If we integrate this expression, we get

$\int \frac{\mathrm{dM}}{M} = \ln M = \text{-} A t + C$, where $C$ is a constant of integration.

At $t = 0 , M = {M}_{i}$, so $C = \ln {M}_{i}$

At any time $t$, $M$ is the mass remaining, ${M}_{R}$.

∴ The integrated rate law in logarithmic form is

$\ln {M}_{R} = \ln {M}_{i} - A t$.

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

$\text{Total energy" = mΔ_cH = 2.6 × 10^6 color(red)(cancel(color(black)("t"))) × (1000 color(red)(cancel(color(black)("kg"))))/(1 color(red)(cancel(color(black)("t")))) × "56 MJ"/(1 color(red)(cancel(color(black)("kg")))) = 1.5 × 10^11color(white)(l) "MJ}$

(d) Mass of ${\text{CO}}_{2}$

${M}_{\textrm{R}} : \textcolor{w h i t e}{m} 16 \textcolor{w h i t e}{m m m m m m l} 44$
$\textcolor{w h i t e}{m m m} \text{CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O}$

${\text{Mass of CO"_2 = 2.6 × 10^6 color(red)(cancel(color(black)("t CH"_4))) × (44 color(white)(l)"t CO"_2)/(16 color(red)(cancel(color(black)("t CH"_4)))) = 7.2 × 10^6 color(white)(l)"t CO}}_{2}$