Assuming Delta_(r )H^(Θ) and Delta_(r ) S^(Θ) to be independent of temperature, at what temperature will the reaction given below becomes spontaneous ?{:(,N_(2)(g),+,O_(2)(g),rarr,2NO(g),,,Delta_(r )H^(Θ)=180.9 kJ mol^(-1)),(S^(Θ)//JK^(-1)mol^(-1),191.4,,204.9,,210.5,,):}

Assuming Delta_(r )H^(Θ) and Delta_(r ) S^(Θ) to be independent of t

1.	Assuming Delta_(r )H^(Θ) and Delta_(r ) S^(Θ) to be independent of temperature, at what temperature will the reaction given below becomes spontaneous ?{:(,N_(2)(g),+,O_(2)(g),rarr,2NO(g),,,Delta_(r )H^(Θ)=180.9 kJ mol^(-1)),(S^(Θ)//JK^(-1)mol^(-1),191.4,,204.9,,210.5,,):}
Answer» Solution :First calculate `Delta_(r )S^(Θ)` for the GIVEN reaction as follows : `Delta_(r )S^(Θ)=sum S_(P)^(Θ)-sum S_(R )^(Θ)=2S_("NO")^(Θ)-S_(N_(2))^(Θ)-S_(O_(2))^(Θ)=2xx210.5-191.4 = 24.7 JK^(-1)MOL^(-1)` Using : `Delta_(r )G^(Θ)=Delta_(r )H^(Θ)-T Delta_(r )S^(Θ)=180.8298xx24.7xx10^(-3)=173.4 kJ mol^(-1)` Clearly, he given endothermic reaction is NON - spontaneous at room temperature. So, we need to increase the temperature to make the reaction spontaneous. `rArr Delta_(r )G^(Θ)=180.8 -(Txx24.7xx10^(-3))kJ mol^(-1)` For SPONTANEITY, put `Delta_(r )G^(Θ)=0` to get : `T_("Switch")=(Delta_(r )H_("system")^(Θ))/(Delta_(r )S_("System")^(Θ))` `rArr T_("switch")=(180.8xx10^(3))/(24.7)~~ 7320 K` The reaction becomes spontaneous above a temperature of 7320 K.