Consider the reaction H2CO(g) + O2(g)CO2(g) + H2O(l) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.63 moles of H2CO(g) react at standard conditions. S°surroundings = J/K

3,116J/K

Explanation:

This question asks to calculate the entropy change of the surroundings.

To do this, we need the standard enthalpy of formation ΔfH° of the reacting species and products first:

We should observe that standard enthalpy if formation of O2 is zero. We proceed with the rest of the species.

H2CO = -109.5KJ/mol

CO2 = -393.5KJ/mol

H2O = -285.8KJ/mol

Now, we calculate the standard change of enthalpy of the reaction as:

ΔHrxn = ΔHproduct - ΔHreactant = (-285.8 - 393.5) +(109.5) = -569.8 KJ/mol

The relationship between the entropy and the standard formation enthalpy is given as

The relationship is:

ΔSosurroundings = - ( ΔHof/ T)

We convert the standard enthalpy of formation to joules first = -569.8 * 10^3 Joules

Using the formula above at a temperature of 298k, the entropy change would be:

-(-569.8 * 10^3)/298 = 1912J/K

Now, we know that 1.63 moles of H2CO reacted. We also need to know the coefficient of the H2CO in the reaction which is 1.

We thus have:

1.63 mol H2CO(g) * (1912J/K * 1 mol H2CO) = 3116J/K

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