Finding Heat of Reaction for MgSO4 and Ba(NO3)2 Solution Mixture

[erik05]

Hello all, this question is really simple but I still seem to get it.

A 50.0 mL sample of 0.500 mol/L MgSO_4 solution at 24.4 degrees celsius is added to 50.0 mL of a 0.500 mol/L Ba(NO_3)_2 solution at 24.4 degrees celsius. The temperature of the mixture rises to 26.3 degrees celsius. Calculate the heat of reaction.

I tried using mc\triangle{t} + mc\triangle{t} but there is no specific heat capacity for either of the compounds. I'm guessing the formula \triangle{H}=nH fits in somewhere but I don't know where. Any hints or suggestions would be appreciated. Thanks.

 

[xanthym]

Hello all, this question is really simple but I still seem to get it.

A 50.0 mL sample of 0.500 mol/L MgSO_4 solution at 24.4 degrees celsius is added to 50.0 mL of a 0.500 mol/L Ba(NO_3)_2 solution at 24.4 degrees celsius. The temperature of the mixture rises to 26.3 degrees celsius. Calculate the heat of reaction.

I tried using mc\triangle{t} + mc\triangle{t} but there is no specific heat capacity for either of the compounds. I'm guessing the formula \triangle{H}=nH fits in somewhere but I don't know where. Any hints or suggestions would be appreciated. Thanks.

The ΔH is determined from the WATER temperature change using {c = 4.184 J/g^oC} for water:
{Water Mass} = (50 g) + (50 g) = (100 g)
{Water ΔT} = (26.3) - (24.4) = (+1.9 ^oC)
{Heat Energy Released} = (100)*(1.9)*(4.184) = (795 J) = (0.795 kJ)

Heat of reaction is generally expressed relative to the primary reactant's mole number, so we use MgSO_4:
{Moles MgSO_4} = (50.0 mL)*(0.500 mol/L) = (0.025 moles)

{Heat of Reaction} = -(0.795 kJ)/(0.025 moles) = (-31.8 kJ/mole)

(Note: Exothermic reactions are by convention expressed with (-)values.)



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[erik05]

Got it. Thanks for the help.

 

[GCT]

You may need to also employ q_{calorimter}. If you were not required to do so, than just ignore.