calculate standard heats THE following REACTION : (H3C)3COH+HBr-----> (H3C)3CBr+H20

Look in your text or notes and find the delta Hf for the compounds listed, then

DHrxn = (n*DHfproducts) -= (n*DHfreactants)

To calculate the standard heat of reaction (∆H°) for the given reaction, we need to know the standard heats of formation (∆H°f) for the reactants and products involved. The standard heats of formation represent the change in enthalpy (∆H) for the formation of 1 mole of a compound from its constituent elements in their standard states at a specified temperature. These values are usually tabulated and can be found in reference sources.

Using the standard heats of formation, we can calculate the standard heat of reaction (∆H°) using the following formula:

∆H° = ∑(∆H°f of products) - ∑(∆H°f of reactants)

Let's find the standard heats of formation for the reactants and products first:

∆H°f[(H3C)3COH] = -476.0 kJ/mol (source: reference table)
∆H°f[HBr] = -36.3 kJ/mol (source: reference table)
∆H°f[(H3C)3CBr] = -70.6 kJ/mol (source: reference table)
∆H°f[H2O] = -285.8 kJ/mol (source: reference table)

Now we can substitute these values into the formula:

∆H° = [∆H°f[(H3C)3CBr] + ∆H°f[H2O]] - [∆H°f[(H3C)3COH] + ∆H°f[HBr]]

∆H° = [-70.6 kJ/mol + (-285.8 kJ/mol)] - [-476.0 kJ/mol + (-36.3 kJ/mol)]

∆H° = -356.4 kJ/mol - (-512.3 kJ/mol)

∆H° = 155.9 kJ/mol

Therefore, the standard heat of reaction (∆H°) for the given reaction is 155.9 kJ/mol.