Calculate Temp Increase for Enthalpy of Sulfur Burning in Excess O2

[MathewsMD]

Homework Statement

A sample of sulfur, mass 2.56g, is burned in excess oxygen inside a calorimeter of heat capacity 923 J/K and insulated by 815g of water. If the sole product of combustion is SO_2(g), what is the expected increase in temperature?

Homework Equations

q = mcΔT
q = CΔT

The Attempt at a Solution

I am given that the ΔH^o_f = -296.8 kJ/mol for SO_2(g) and the enthalpy of formation of oxygen gas and sulfur is 0.

S_(s) + O_2(g) → SO_2(g)

So, I did:

(1g / 32.06 g/mol)(-296.8 kJ/mol) = - [(4.184 g/molK)(815g)ΔT + (923 J/K)(ΔT)]

After isolating for ΔT and converting units to J, i found my answer as 2.136 K but this is incorrect. Any comments on where I went wrong? Thanks!

 

[Chestermiller]

Homework Statement

A sample of sulfur, mass 2.56g, is burned in excess oxygen inside a calorimeter of heat capacity 923 J/K and insulated by 815g of water. If the sole product of combustion is SO_2(g), what is the expected increase in temperature?

Homework Equations

q = mcΔT
q = CΔT

The Attempt at a Solution

I am given that the ΔH^o_f = -296.8 kJ/mol for SO_2(g) and the enthalpy of formation of oxygen gas and sulfur is 0.

S_(s) + O_2(g) → SO_2(g)

So, I did:

(1g / 32.06 g/mol)(-296.8 kJ/mol) = - [(4.184 g/molK)(815g)ΔT + (923 J/K)(ΔT)]

After isolating for ΔT and converting units to J, i found my answer as 2.136 K but this is incorrect. Any comments on where I went wrong? Thanks!

A negative heat of reaction means that, to maintain the temperature constant at the initial temperature, you need to remove heat. So the reaction gives off 296.8 kJ/mol. So the temperature in your calorimeter has to rise. There shouldn't be a minus sign on the rhs.