Question

Consider the reaction C2H4(g) + H2O(g)CH3CH2OH(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of C2H4(g) react at standard conditions. S°surroundings = J/K

Answer 1

Answer: The value of $\Delta S^o$ for the surrounding when given amount of ethene gas is reacted is 238.80 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$C_2H_4(g)+H_2O(g)\rightarrow CH_3CH_2OH(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(CH_3CH_2OH(g))})]-[(1\times \Delta S^o_{(C_2H_4(g))})+(1\times \Delta S^o_{(H_2O(g))})]$

We are given:

$\Delta S^o_{(CH_3CH_2OH(g))}=282.7J/K.mol\\\Delta S^o_{(C_2H_4(g))}=219.56J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times (282.7))]-[(1\times (219.56))+(1\times (188.82))]\\\\\Delta S^o_{rxn}=-125.68J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-125.68) J/K = 125.68 J/K

We are given:

Moles of ethene gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of ethene gas is reacted, the entropy change of the surrounding will be 125.68 J/K

So, when 1.90 moles of ethene gas is reacted, the entropy change of the surrounding will be = $\frac{125.68}{1}\times 1.90=238.80J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of ethene gas is reacted is 238.80 J/K