At `25^(@)C , Delta H^(0)` for the reaction , `N_(2)(g)+(1)/(2)O_(2)(g) to N_(2)O(g)`, is `+82 kJ`. At this temperature if the standard molar entropies for `N_(2)(g),O_(2)(g) and N_(2)O(g)` are 191.6, 205.2 and `219.9 J. K^(-1). " mol"^(-1)` respectively, calculate `Delta S_("univ")^(0)` for the reaction.

Explanation

For the given reaction, the standard molar entropy change,
`Deltaoverline(S)_(N_(2)O(g))^(0)-(overline(S)_(N_(2)(g))^(0)+(1)/(2)overline(S)_(O_(2)(g))^(0)) [ overline(S)^(0)`= standard molar entropy]
`= [ 219.9-(191.6+(1)/(2)205.2)]J.K^(-1)= - 74.3 J. K^(-1)`
Entropy change for the surroundings, `Delta S_("surr")^(0)= - (Delta H^(0))/(T)`
`= (82xx10^(3))/((273+25)K)= - 275.2 J. K^(-1)`
`therefore Delta S_("univ")^(0)= Delta S_("sys")^(0) + DeltaS_("surr")^(0) = (-74.3-275.2) J. K^(-1)`
`= -349.5 J. K^(-1)`