Question

Consider the reaction: H2(g) + Cl2(g) --> 2HCl(g)

Answer 1

Answer : The entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition is 45.8 J/K

Explanation :

The given balanced reaction is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

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

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

$\Delta S^o=[n_{HCl}\times \Delta S_f^0_{(HCl)}]-[n_{H_2}\times \Delta S_f^0_{(H_2)}+n_{Cl_2}\times \Delta S_f^0_{(Cl_2)}]$

where,

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

n = number of moles

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

$\Delta S_f^0_{(H_2)}$ = 130.684 J/mol.K

$\Delta S_f^0_{(Cl_2)}$ = 223.066 J/mol.K

$\Delta S_f^0_{(HCl)}$ = 186.908 J/mol.K

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

$\Delta S^o=[2mole\times (186.908J/K.mole)]-[1mole\times (130.684J/K.mole)+1mole\times (223.066J/K.mole)}]$

$\Delta S^o=20.066J/K$

Now we have to calculate the entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition.

From the reaction we conclude that,

As, 1 moles of $H_2$ has entropy change = 20.066 J/K

So, 2.28 moles of $H_2$ has entropy change = $\frac{2.28}{1}\times 20.066=45.8J/K$

Therefore, the entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition is 45.8 J/K