Is the reaction between sodium sulfate and strontium nitrate endothermic or exothermic and why?
To determine whether the reaction between sodium sulfate (Na2SO4) and strontium nitrate (Sr(NO3)2) is endothermic or exothermic, we need to examine the enthalpy of the reaction. The enthalpy change (∆H) represents the heat energy released or absorbed during a chemical reaction.
To find the value of ∆H for the reaction, we can look up the enthalpies of formation (∆Hf) for each compound involved in the reaction. The enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its constituent elements, with all species in their standard state (usually 298 K and 1 atm).
The balanced chemical equation for the reaction is as follows:
Na2SO4 + Sr(NO3)2 → 2NaNO3 + SrSO4
Now, let's look up the enthalpies of formation for the compounds involved:
∆Hf(NaNO3) = -467.7 kJ/mol
∆Hf(SrSO4) = -1467.8 kJ/mol
∆Hf(Na2SO4) = -1387.6 kJ/mol
∆Hf(Sr(NO3)2) = -1110.6 kJ/mol
To calculate the overall enthalpy change, we sum the enthalpies of formation of the products and subtract the sum of the enthalpies of formation of the reactants:
∆H = [2∆Hf(NaNO3) + ∆Hf(SrSO4)] - [∆Hf(Na2SO4) + ∆Hf(Sr(NO3)2)]
Substituting the values, we get:
∆H = [2(-467.7 kJ/mol) + (-1467.8 kJ/mol)] - [(-1387.6 kJ/mol) + (-1110.6 kJ/mol)]
After evaluating the expression, we find that the value of ∆H is negative, indicating that the reaction is exothermic. The negative value of ∆H means that the reaction releases heat energy to the surroundings.
Therefore, the reaction between sodium sulfate and strontium nitrate is exothermic because it releases heat energy during the reaction.
dHfreaction = (n*dHf products) - (n*dH products)
Look up the numbers ad calculate.