Given the following entropy values (in `JK^(-1)"mol"^(-1))` at 298 K atm : `H_(2)(g):130.6.Cl_(2)(g):223.0andHCl(g): 186.7` . The entropy change (in `JK^(-1)"mol"^(-1))` for the reaction
`H_(2)(g)+Cl_(2)(g)to2HCl(g)`is

Options

A. `+540.3`
B. `+727.0`
C. `-166.9`
D. 19.8

Answer

A

Explanation

`DeltaS^(o)=sumS_((p))^(o)-sumS_((R))^(o)`
`Delta S^0 = 2S_("HCl")^0-(S_(H_2)^0 +S_(Cl_2)^0)`
`= 2 xx 186.7-(130.6+232.0)=19.8 kJ mol^(-1)`