Question

Calculate the standard entropy change for the reaction H2+Cl2=2HCl at 25 degree centigrade given delta s0 for H2 & Cl2 & HCl are 0.13 , 0.22 , 0.19 Kj/K mol^-1 respectively

Answer 1

∆S° = 1mole of H2 + 1 mole of cl2 - 2 mole of HCL.

0.13+ 0.22 - 2×0.19
∆S°= -0.03

TY MATE

Answer 2

Answer: The value of $\Delta S^o$ for the reaction is 0.03 kJ/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:

$H_2+Cl_2\rightarrow 2HCl$

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HCl)})]-[(1\times \Delta S^o_{(H_2)})+(1\times \Delta S^o_{(Cl_2)})]$

We are given:

$\Delta S^o_{(HCl)}=0.19kJ/K.mol\\\Delta S^o_{(Cl_2)}=0.22kJ/K.mol\\\Delta S^o_{(H_2)}=0.13kJ/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (0.19))]-[(1\times (0.22))+(1\times (0.13))]\\\\\Delta S^o_{rxn}=0.03kJ/K$

Hence, the value of $\Delta S^o$ for the reaction is 0.03 kJ/K