Question

If 1.6g of CH4 reacts with oxygen gas to form water and carbon dioxide what is the change in entropy for the universe?

I don't get how to calculate the entropy, the part where s= .1(0.2) ext.....

Answer 1

The reaction we had is CH4 (g) + 2O2(g) --> CO2 (g) + 2H2O(g)

dHo reaction = dHo ( CO2) + 2dHo ( H2O (g)) - dHo ( CH4) - 2dHo ( O2) ( The values of enthalpies dHo taken from web)

= ( - 393.5 KJ/mol) + ( - 242 KJ/mol) - ( - 74.9 KJ?mol) - 2 ( 0)

= - 560.6 KJ/mol

CH4 moles reacted = ( mass / Molar mass of CH4) = ( 1.6 g / 16g/mol) = 0.1 mol

Thus heat released in reaction = ( moles of CH4) x enthalphy of reaction = 0.1 mol x ( - 560.6 KJ/mol)

= - 56.06 KJ = - 56060 J ( -ve sign indicates heat released)

Entropy universe = - ( dH of reaction / T)

= - ( - 56060 J ) / ( 298 K) = 1881 J/K