Question

If S° for H2,cl2 and hcl are 0.13 ,0.22 and 0.19KJ K mol respectively.the total change in standard entropy for the reaction. H2+Cl2-------2HCL

Answer 1

here is your answer mate

Answer 2

Given:

S° for $H_2$ = 0.13 $J K^{-1} mol^{-1}$

S° for $Cl_2$ = 0.22 $J K^{-1} mol^{-1}$

S° for HCl = 0.19 $J K^{-1} mol^{-1}$

To find:

Total change in standard entropy = ?

Formula to be used:

$\Delta S_{\text {Reaction }}^{\circ}=\Delta S_{\text {product }}^{\circ}-\Delta S_{\text {Reactant }}^{\circ}$

Calculation:

As the $\Delta S}^{\circ$ values of product and reactant are given, the change in standard entropy is calculated through the following steps:

In the following given reaction is:

$H_2+Cl_2\rightarrow 2HCl$

Reactants are $H_2$ and $Cl_2$, whereas product is 2HCl

∴$\Delta S_{\text {Reaction }}^{\circ}=\Delta S_{\text {product }}^{\circ}-\Delta S_{\text {Reactant }}^{\circ}$

$\Delta S_{\text {Reaction }}^{\circ}= 2 (0.19) - (0.13+0.22)$

$\Delta S_{\text {Reaction }}^{\circ}= 0.38 - 0.35$

$\Delta S_{\text {Reaction }}^{\circ}= 0.03$

Conclusion:

Thus the total change in standard entropy for the reaction is 0.03 $J K^{-1} mol^{-1}$.

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