A sample of octane burns releasing 2290 J of heat to the surroundings, and the gases produced expands against a piston to do 560 joules of work. Calculate the internal energy change for this reaction.
1- –2850 J
2- –1730 J
3- –2290 J
4- +1730 J
5- +2850 J

The internal energy change for the reaction is -2850 J

Explanation:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

Work done for expansion process is taken as negative and work done for compression is taken as positive.

According to the First law of thermodynamics,

where,

= internal energy

q = heat absorbed or released  = -2290 J

w = work done = -560 J

Putting values in above equation, we get:

Hence, the internal energy change for the reaction is -2850 J

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.