Question

For a reaction ,$2X(s) + 2Y(s) \to 2C(l) + D(g)$ , $\Delta H$ at ${27^ \circ }C$ is $- 28Kcalmo{l^{ - 1}}$.If $\Delta U$ is $X$ $Kcalmo{l^{ - 1}}$. Find $X$.
A.$X = + 25.5$
B.$X = - 28.6$
C. $X = - 25.5$
D.$X = 28.4$

Answer 1

The measurement of energy in a thermodynamic system is enthalpy. The overall content of heat in a system is enthalpy, which is equal to the system's internal energy plus the product of volume and pressure. In a chemical process at constant pressure, an enthalpy change is defined as the difference between the energy acquired by the production of new chemical bonds and the energy used to break bonds.
Formula Used:
$\Delta U = \Delta H - \Delta nRT$
$\Delta U =$Change in internal energy
$\Delta H =$Change in enthalpy
$\Delta n =$Change in moles of gaseous molecules
$R =$Gas constant $= 2k/calmo{l^{ - 1}}$
$T =$ Temperature in Kelvin

Complete answer:
Given:
$T = 27 + 273 = 300K$
$\Delta H = - 28Kcalmo{l^{ - 1}}$
$\Delta n = 1$
To find:
$\Delta U =$ ?
Using the above formula,
$\Delta U = \Delta H - \Delta nRT$
Substituting the values in the above equation,
$\Delta U = - 28 - (1)(2 \times {10^{ - 3}}) \times 300$
$\Delta U = - 28 - 0.6$
$\Delta U \Rightarrow - 28.6Kcal/mol$
Hence, the correct option is B.$X = - 28.6$.

Note:
Internal energy is the total amount of energy stored in the device. It is the quantity of potential and kinetic energy stored by the mechanism. The amount of internal energy plus the combination of the system's gas pressure and length is referred to as enthalpy. The heat energy produced is often used in thermodynamics to improve the system’s energy or to do some useful work. The energy associated with an open system is called enthalpy, which is often greater than or equal to internal energy.