Question

What is ΔS° for the reaction SO3(g) + H2O(l) → H2SO4(aq)? Substance S°(J/K • mol) SO3(g) 256.2 H2O(l) 69.9 H2SO4(aq) 156.9

Answer 1

Answer : The value of $\Delta S^o$ for the reaction is, -169.2 J/K

Explanation :

The given balanced reaction is,

$SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2SO_4}\times \Delta S_f^0_{(H_2SO_4)}]-[n_{SO_3}\times \Delta S_f^0_{(SO_3)}+n_{H_2O}\times \Delta S_f^0_{(H_2O)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(H_2SO_4)}$ = 156.9 J/mol.K

$\Delta S_f^0_{(SO_3)}$ = 256.2 J/mol.K

$\Delta S_f^0_{(H_2O)}$ = 69.9 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (156.9J/K.mole)]-[1mole\times (256.2J/K.mole)+1mole\times (69.9J/K.mole)}]$

$\Delta S^o=-169.2J/K$

Therefore, the value of $\Delta S^o$ for the reaction is, -169.2 J/K

Answer 2

The standard entropy change for the reaction SO₃(g) + H₂O(l) → H₂SO₄(aq) is -169.2 J/K.

Let's consider the following balanced equation.

SO₃(g) + H₂O(l) → H₂SO₄(aq)

Given the standard entropies (S°) of reactants (r) and products (p), we can calculate the standard entropy change for the reaction (ΔS°) using the following expression.

$\Delta S\° = \Sigma S\°(p) \times n_p - \Sigma S\°(r) \times n_r\\\\\Delta S\° = S\°(H_2SO_4(aq)) \times 1 mol - S\°(SO_3(g)) \times 1 mol - S\°(H_2O(l)) \times 1 mol\\\\\Delta S\° = (156.9 J/K.mol) \times 1 mol - (256.2 J/K.mol) \times 1 mol - (69.9J/K.mol) \times 1 mol = -169.2J/K$

where,

• n are the number of moles

The standard entropy change for the reaction SO₃(g) + H₂O(l) → H₂SO₄(aq) is -169.2 J/K.

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