In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surroundings is 1.1 MJ, and the work done on the system is 13.2 kcal. Calculate ΔE (in kJ). Be sure to include the correct sign (+/-). Enter to 0 decimal places.

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, distinguishing two kinds of transfer of energy, as heat and as thermodynamic

The formula used for the calculation is as follows:-

The value of q and w is given in the question, after applying the equation.

After converting, the value of W into joules is:-

Then the value of E is

Hence, the correction energy is 1155.2j

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is 1155 kJ

Explanation:

According to first law of thermodynamics:

=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=  {Work done on the system is positive as the final volume is lesser than initial volume}

w =   (1kcal=4.184 kJ)

q = +1.1 MJ = 1100 kJ  (1MJ=1000kJ)   {Heat absorbed by the system is positive}

Thus is 1155 kJ

im pretty sure its b

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