Question

The reaction of nitrogen with hydrogen to make ammonia has \[\Delta H = - 92Kj\] ${N_2}_{(g)} + 3{H_{2(g)}} \to 2N{H_3}_{(g)}$ . What is the value of $\Delta U(in{\text{ }}Kj)$ if the reaction is carried out at a constant pressure of $40bar$ and the volume change is $ - 1.25litre$ ?

Answer 1

Hint: The absorption or loss of heat from a system is known as its enthalpy of system or enthalpy of reaction if the reaction is occurring. And the change in internal energy is referred to the change in the energy of particles present in the system.

Complete step by step answer:
The branch of physics that deals with heat and transfer of energy is known as thermodynamics. It is also used in chemistry and is known as chemical thermodynamics. It basically deals with interrelation of heat and work and the physical changes noted with the help of laws of thermodynamics.
The three laws related to chemical thermodynamics are as follows:
The first law of thermodynamics is the Law of conservation of energy, energy can neither be created nor be destroyed. The second law of thermodynamics states that the entropy of an isolated system keeps on increasing. The third law states that as temperature approaches to absolute zero the entropy value for the system becomes constant.
The internal energy of a thermodynamic system is represented as \[\Delta U\] .
The amount of heat absorbed or lost from the system at constant pressure is termed as enthalpy and it is represented as \[\Delta H\].
The relation between pressure, \[\Delta U\] and \[\Delta H\] is as follows:
\[\Delta H = \Delta U + P\Delta V\] where, ${\text{P and }}\Delta {\text{V}}$ are the pressure and the volume change of the system respectively.
We have a system in which nitrogen and hydrogen come together to produce ammonia.
For this system the provided data is as follows:
 \[\Delta H = - 92Kj\]
$P = 40bar$
\[\Delta V = - 1.25L\]
The negative values indicate the decrease in the respective chemical terms.
So we will put all the values in formula and find the unknown $\Delta U$ .
\[\Delta H = \Delta U + P\Delta V\]
\[\Delta U = \Delta H - P\Delta V\]
Convert the energy in joules and pressure in atm.
\[\Delta U = - 92 \times {10^3} - (40 \times 1.013)( - 1.25)\] ($1atm = 1.013bar$)
By calculating we get,
$\Delta U = - 91.949 \times {10^3}j$
$\Delta U = - 91.949Kj$
Hence the change in internal energy is $ - 91.949Kj$.

Note:
As we have seen in our calculations that some of our values are negative, it doesn’t mean that they are actually negative, it depicts the loss. Like, change in internal energy has come in negative so it shows that the internal energy is decreased from the system.