Question

In a particular reaction 2 kJ

of heat is released by the system and 6 kJ

of work is done on the system. Determine

of ∆H and ∆U?​

Answer 1

$\huge\underline{\underline\mathtt{Answer:-}}$

Solution : According to the first law of

thermodynamics

∆U = Q + W

Q = -2 kJ, W = +6 kJ

∆U = -2 kJ + 6 kJ = + 4 kJ

Qp = ∆H = - 2kJ

Answer 2

The correct answer to the above question is given below -

Given: Heat released = 2 kJ

           Work done on the system = 6 kJ

To find: Determine ΔH and ΔU

Solution:

According to the first law of thermodynamics - "Energy can neither be created nor be destroyed, it can only be transformed from one source of energy to another."

ΔU = Q + W (Where ΔU is the change in internal energy, Q is the heat energy absorbed and W is the amount of work done.)

Since heat is released, therefore, Q = -2 kJ, and Work is done on the system so W = +6 kJ

ΔU = -2 + 6

     = 4 kJ

Since ΔH is the amount of heat absorbed or released by the system, therefore, ΔH will be equal to - 2kJ

ΔU = +4 kJ

ΔH = -2 kJ