# Methane, CH_4, reacts with oxygen to produce carbon dioxide, water, and heat. CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l); DeltaH = -890.3 (kJ)/(mol). What is the value of DeltaH if 7.00 g of CH_4 is combusted?

Jul 10, 2016

$\Delta H = - \text{389 kJ}$

#### Explanation:

The problem provides you with the thermochemical equation that describes the combustion of methane, ${\text{CH}}_{4}$

${\text{CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))" "DeltaH = -"890.3 kJ mol}}^{- 1}$

Notice that the thermochemical equation includes the enthalpy change of combustion, which as you can see is given in kilojoules per mole, ${\text{kJ mol}}^{- 1}$.

This tells you that when one mole of methane undergoes combustion, $\text{890.3 kJ}$ of heat are given off. This is equivalent to saying that the enthalpy change for the combustion of one mole of methane is

$\Delta H = - \text{890.3 kJ}$

The minus sign is used to denote heat given off.

So, you know the enthalpy change of reaction associated with the combustion of one mole of methane. Use methane's molar mass to calculate how many moles you have in your sample

7.00 color(red)(cancel(color(black)("g"))) 8 "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "0.4364 moles CH"_4

Use the enthalpy change of combustion to find the heat given off when $0.4364$ moles of methane undergo combustion

0.4364 color(red)(cancel(color(black)("moles CH"_4))) * "890.3 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "389 kJ"

Now, the enthalpy change of reaction, $\Delta H$, will be

$\Delta H = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{- \text{389 kJ}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Once again, keep in mind that the minus sign is needed because the heat is being given off.

The answer is rounded to three sig figs.