Calculate the equilibriumconstant for the following reaction at 298 K and 1 atm pressure : NO(g) + (1)/(2) O_(2)(g) hArr NO_(2)(g) Given : Delta_(f) H^(@) at 298K For NO(g) = 90.4 kJ mol^(-1),ForNO_(2)(g) = 33.8 kJ mol^(-1) Delta S^(@)at 298 K for the reaction= - 70.8 J K^(-1)mol^(-1) Gas constant, R= 8.31 JK^(-1) mol^(-1)

Calculate the equilibriumconstant for the following reaction at 298 K

1.	Calculate the equilibriumconstant for the following reaction at 298 K and 1 atm pressure : NO(g) + (1)/(2) O_(2)(g) hArr NO_(2)(g) Given : Delta_(f) H^(@) at 298K For NO(g) = 90.4 kJ mol^(-1),ForNO_(2)(g) = 33.8 kJ mol^(-1) Delta S^(@)at 298 K for the reaction= - 70.8 J K^(-1)mol^(-1) Gas constant, R= 8.31 JK^(-1) mol^(-1)
Answer» Solution :`Delta_(r)H^(@) = Delta_(f) H^(@) ( NO_(2)) -[Delta_(f) H^(@) (NO)+(1)/(2) Delta_(f) H^(@) (O_(2))]= 33.8 -[90.4 + 0] = - 56.6kJ mol^(-1)` `= - 56600 J mol^(-1) - 298 K ( - 70.8 JK^(-1) mol^(-1)) = - 35501.6 J mol^(-1)` `:. - 35501.6 =- 2.303 xx 8.31 xx298 log K ` or `log K = 6.2250` or `K = 1.679 xx 10^(6)`