Question

Consider the reaction CH4(g) 2O2(g)CO2(g) 2H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions.

Answer 1

Answer:

the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K

Explanation:

The balanced chemical equation of the reaction in the question given is:

$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2 H_2O _{(g)}$

Using standard thermodynamic data at 298K.

The entropy of each compound above are listed as follows in a respective order.

Entropy of (CH4(g)) = 186.264 J/mol.K

Entropy of (O2(g)) = 205.138 J/mol.K

Entropy of (CO2(g)) = 213.74 J/mol.K

Entropy of (H2O(g)) = 188.825 J/mol.K

The change in Entropy (S) of the reaction is therefore calculated as follows:

$=1*S(CO2(g)) + 2*S(H2O(g)) - 1*S( CH4(g)) - 2*S(O2(g))$

$=1*(213.74) + 2*(188.825) - 1*(186.264) - 2*(205.138)$

=  -5.15  J/mol.K

Given that :

the number of moles = 1.62 of CH4(g) react at standard conditions.

Then;

The change in entropy of the rxn $= 1.62 \ mol * -5.15 \ J/mol.K$

= −8.343 J/K