# Tutorial 5 - 2010 - Solution

The heat of reaction for a certain reaction is given by $$\Delta H_{r}^{0}(T)= -24097 -0.26 T+1.69\times 10^{-3}T^2 + {\displaystyle{\frac{1.5\times10^5}{T}}}\;$$ cal/mol. Compute the temperature at which $$\Delta H_{r}^{0}(T)= -23505$$ cal/mol.