Question

Click on the Hess's law of constant heat summation button within the activity and use the example shown to calculate the reaction enthalpy, ΔH, for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)Use the series of reactions that follow: C(s)+2H2(g)→CH4(g), ΔH =−74.8 kJ. C(s)+O2(g)→CO2(g), ΔH =−393.5 kJ. 2H2(g)+O2(g)→2H2O(g), ΔH =−484.0 kJ. H2O(l)→H2O(g), ΔH =44.0 kJ.

Answer 1

Answer: The reaction enthalpy of given reaction is -890.7KJ.

Solution:

Given:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l),\Delta H= ?$...(1)

$C(s)+2H_2(g)\rightarrow CH_4(g),\Delta H=-74.8KJ$...(2)

$C(s)+O_2(g)\rightarrow CO_2(g), \Delta H=-393.5KJ$...(3)

$2H_2(g)+O_2(g)\rightarrow2H_2O(g), \Delta H=-484.0KJ$...(4)

$H_2O(l)\rightarrow H_2O(g),\Delta H=44.0KJ$...(5)

According to Hess law: "The enthalpy of the chemical reaction remains the same whether reaction is carried out in single step or in multiple steps."

$\Delta H\text{of equation}(1)=(3)+(4)-(2)-2\times (5)$

$\Delta H\text{of equation}(1)=(-393.5)+(-484.0)-(-74.8)-2\times (44.0)=-890.7KJ$

The reaction enthalpy of given reaction:

$CH_4(g)+2O_2(g)\rightarrowCO_2(g)+2H_2O(l),\Delta H=-890.7KJ$

Answer 2

It is a little hard to explain step-by-step what to do so I hope you can follow what to do by looking at how I arranged the equations.

2 H2O(g) --------------------------> 2 H2O(l) -88
2 H2 + O2 ----------------------------> 2 H2O(g) -484
C + O2 ----------------------------> CO2 -393.5
CH4 -------------------------------> C + 2 H2 + 74.8
________________________________________...
CH4 + 2 O2 ---------------------> CO2 + 2 H2O -890.6 kJ

Once you cancel out the substances that appear BOTH on the left AND the right of the arrows you are left with your net equation.