Calculate the standard Gibbs energy change for the formation of propane at 298 K 3C ( graphite ) + 4H_(2)(g) rarr C_(3)H_(8)(g) Delta_(f)H^(@)for propane, C_(3)H_(8) (g),is - 103.8 kJ mol^(-1) Given S_(m)^(@) C_(3)H_(8)(g) = 270.2JK^(-1) mol^(-1), S_(m)^(@) C( graphite )= 5.70 JK^(-1) mol^(-1) and S_(m)^(@) H_(2)(g) = 130 .7 JK^(-1) mol^(-1)

Calculate the standard Gibbs energy change for the formation of propan

1.	Calculate the standard Gibbs energy change for the formation of propane at 298 K 3C ( graphite ) + 4H_(2)(g) rarr C_(3)H_(8)(g) Delta_(f)H^(@)for propane, C_(3)H_(8) (g),is - 103.8 kJ mol^(-1) Given S_(m)^(@) C_(3)H_(8)(g) = 270.2JK^(-1) mol^(-1), S_(m)^(@) C( graphite )= 5.70 JK^(-1) mol^(-1) and S_(m)^(@) H_(2)(g) = 130 .7 JK^(-1) mol^(-1)
Answer» SOLUTION :`Delta_( r)S^(@) = S^(@) ( C_(3)H_(8)) -[3S^(@) ` ( graphite) ` + 4S^(@) ( H_(2))]= 270.2- ( 3 xx 5.7 xx 4xx 130.7 ) = - 269.7 JK^(-1) MOL^(-1)` `Delta_(r)G^(@) = Delta_(r)H^(@) - T Delta_(r)S^(@) = - 103.8 kJ mol^(-1) - 298 K xx ((-269.7)/( 1000) kJ K^(-1) mol^(-1))` `= - 103.8 + 80.37 kJ mol^(-1) = - 23.43 kJ mol^(-1)` .