Question

Calculate δh o for the reaction. enter your answer in the provided box. ch3oh + hbr → ch3br + h2o

Answer 1

To calculate the enthalpy change for a reaction, the enthalpy changes of the individual reactions involved are needed.

Step-by-step explanation:

The enthalpy change, ΔHo, for a reaction can be calculated using Hess's Law. To calculate ΔHo for the reaction CH3OH + HBr → CH3Br + H2O, we need to use the enthalpy changes of the individual reactions involved. The given information does not include the enthalpy changes for the reactions involved, so we cannot calculate ΔHo without that information.

Answer 2

The standard enthalpy change (ΔH°) for the given reaction is approximately -103.8 kJ/mol.

To calculate the standard enthalpy change (ΔH°) for the given reaction:

CH3OH + HBr → CH3Br + H2O

You can use the standard enthalpy of formation values for each compound involved in the reaction. The standard enthalpy of formation (ΔHf°) represents the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

Here are the standard enthalpy of formation values:

- ΔHf° for CH3OH (methanol) = -238.6 kJ/mol

- ΔHf° for HBr (hydrogen bromide) = -36.4 kJ/mol

- ΔHf° for CH3Br (methyl bromide) = -93.0 kJ/mol

- ΔHf° for H2O (water) = -285.8 kJ/mol

Now, you can use these values to calculate ΔH° for the reaction:

ΔH° = ΣΔHf°(products) - ΣΔHf°(reactants)

ΔH° = [ΔHf°(CH3Br) + ΔHf°(H2O)] - [ΔHf°(CH3OH) + ΔHf°(HBr)]

ΔH° = [(-93.0 kJ/mol) + (-285.8 kJ/mol)] - [(-238.6 kJ/mol) + (-36.4 kJ/mol)]

Now, calculate the values:

ΔH° = [-378.8 kJ/mol] - [-275.0 kJ/mol]

ΔH° = -378.8 kJ/mol + 275.0 kJ/mol

ΔH° = -103.8 kJ/mol

So, the standard enthalpy change (ΔH°) for the given reaction is approximately -103.8 kJ/mol.