Consider the following standard heats of formation:
p₄o₁₀(s) = -3110 kj/mol
h₂o(l) = -286 kj/mol
h₃po₄(s) = -1279 kj/mol
calculate the change in enthalpy for the following process:

The Standard enthalpy of reaction:            

Explanation:

Given- Standard Heat of Formation:

= -3110 kJ/mol,

= -286 kJ/mol,

= -1279 kJ/mol

Given chemical reaction: P₄O₁₀(s) + 6H₂O → 4H₃PO₄

The standard enthalpy of reaction: = ?

To calculate the Standard enthalpy of reaction (), we use the equation:

Therefore, the Standard enthalpy of reaction:

since electrons are negative the answer is neagtaive