Standard enthelpy of formation

[krootox217]

Homework Statement

I have the following task:

The Standard enthalpy of formation of gaseous H2O at 298K is -241.82 kJ mol-1. Estimate its value at 100 °C given the following values of the molar heat capacities at constant pressure: H2O (g): 33.58 JK-1mol-1, H2 (g): 28.84 JK-1mol-1, O2 (g): 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature.

Homework Equations

The Attempt at a Solution

The only formula I can think of is this one:

But I don't know how to calculate this exactly. Can some help me getting started?

 

[insightful]

First, you must devise a pathway to get from reactants at 100C, through the reaction at 25C, and finally to the product at 100C.

 

[krootox217]

Ok, thx

is it correct that the standard enthalpie of H2 and O2 are 0 because of the definition?

Does this mean I that I can use my formula in this way?

 

[insightful]

Ok, thx

is it correct that the standard enthalpie of H2 and O2 are 0 because of the definition?

Does this mean I that I can use my formula in this way?

All correct.

 

[krootox217]

Therefore the ΔH(T1) is -241.82 kJ mol-1, the Cp(H2O) is 33.58 JK-1mol-1, the Cp(H2) is 28.84 JK-1mol-1, the Cp(O2) is 29.37 JK-1mol-1 and I just need to insert these values in the equation above and I get the result for ΔH at 373K and therefore my final result?

 

[insightful]

Therefore the ΔH(T1) is -241.82 kJ mol-1, the Cp(H2O) is 33.58 JK-1mol-1, the Cp(H2) is 28.84 JK-1mol-1, the Cp(O2) is 29.37 JK-1mol-1 and I just need to insert these values in the equation above and I get the result for ΔH at 373K and therefore my final result?

Check all your units for consistency.

 

[krootox217]

I think everything is fine, I just need to transform ΔH(T1) from -241.82 kJ mol-1 to -241'820 Jmol-1 and then it should be consistent, I think

 

[insightful]

Go for it! (Note: If you had left it as kJ, you'd have gotten a much different answer. Experience will tell you that heats of reaction don't change drastically with moderate temperature change, like here, so a "wild" answer is a tipoff that something's wrong.)

 

[krootox217]

Thanks a lot!