Chapter 5 - Thermochemistry - Exercises - Page 206: 5.46a

$\Delta H = -40.528kJ$

We can use a simple proportion calc to find the answer: If the formation of 3 moles of $O_2$ has a $\Delta H = -89.4kJ$, 1.36 moles will have: $\frac{3}{-89.4} = \frac{1.36}{x}$ $3x = 1.36 * - 89.4$ $3x = -121.584$ $x = -40.528kJ$