## Chemistry: The Central Science (13th Edition)

$\Delta H^o_f : C_4H_{10}O = -279.45kJ/mol$
As we found in 3.78a: $C_4H_{10}O(l) + 6O_2(g) --> 4CO_2(g) + 5H_2O(l)$ $\Delta H^o : -2723.7 kJ$ $\Delta H^o_f : O_2(g) = 0 kJ/mol$ $\Delta H^o_f : CO_2(g) = -393.5 kJ/mol$ $\Delta H^o_f : H_2O(l) = -285.83 kJ/mol$ $\Delta H^o_f : C_4H_{10}O(l) = x$ *These values are given in the book $-2723.7 = [ 4*(-393.5) + 5*(-285.83) ] - [(x) + 6*(0) ]$ $-2723.7 = [-1574 + (-1429.15)] - x$ $-2723.7 = -3003.15 - x$ $279.45 = -x$ $x = -279.45$