Calculate Delta_(r)G^(0) for (NH_(4)Cl, s) at 310K. Given: Dleta _(f)H^(0) for (NH_(4)Cl, s) = - 314.5 KJ/mol, Delta_(r ) C_(P) = 0S_(N_(2)(g))^(0) = 192JK^(-1) mol^(-1), S_(H_(2) (g))^(0) = 130.5 JK^(-1) mol^(-1), S_(Cl_(2)(g))^(0) = 233JK^(-1) mol^(-1), S_(NH_(4)Cl(s))^(0) = 99.5 JK^(-1) mol^(-1). All given data at 300K

Calculate Delta_(r)G^(0) for (NH_(4)Cl, s) at 310K. Given: Dleta _(f)H

1.	Calculate Delta_(r)G^(0) for (NH_(4)Cl, s) at 310K. Given: Dleta _(f)H^(0) for (NH_(4)Cl, s) = - 314.5 KJ/mol, Delta_(r ) C_(P) = 0S_(N_(2)(g))^(0) = 192JK^(-1) mol^(-1), S_(H_(2) (g))^(0) = 130.5 JK^(-1) mol^(-1), S_(Cl_(2)(g))^(0) = 233JK^(-1) mol^(-1), S_(NH_(4)Cl(s))^(0) = 99.5 JK^(-1) mol^(-1). All given data at 300K
Answer» `-198.56kJ//mol` `-426.7 kJ//mol` `2-2.3 kJ//mol` none of these Solution :`(1)/(2) N_(2) + 2H_(2)+ (1)/(2) Cl_(2) rarr NH_(4) Cl` `Delta G_(gamma)^(0) = Delta H_(gamma)^(0) - T Delta S_(gamma)^(0)` `Delta H_(310)^(0) = Delta H_(300)^(0) = - 314.5` `Delta S_(gamma)^(0) = 99.5 - ((1)/(2) xx 192 + 2 xx 130.5 + (1)/(2) xx 233)` `= - 374` `Delta G^(0) = - 314.5 - (310 xx (-374))/(1000)` `= - 198.56` Kj/mol