InterviewSolution
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InterviewSolution
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Calculate the equilibrium constant 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), ` For `NO_(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)` |
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Answer» Correct Answer - `1.679xx 10^(6)` `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)` |
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