Calculate the internal energy for the reaction shown below which was performed at constant pressure of 40 Pa and a volume of 1.0 m3.
N2 +3H2→ 2 NH3 ΔH = -899 J
The internal energy, ΔU, can be calculated using the following equation:
ΔU = ΔH - PΔV
where ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in volume. At constant pressure, ΔV is equal to the initial volume, which is 1.0 m3. Plugging in the values:
ΔU = -899 J - (40 Pa)(1.0 m3)
ΔU = -899 J - 40 J
ΔU = -939 J
Therefore, the internal energy change for the reaction is -939 J.
To calculate the internal energy, we can use the equation:
ΔU = ΔH - PΔV
where:
ΔU is the change in internal energy,
ΔH is the change in enthalpy,
P is the pressure, and
ΔV is the change in volume.
Given that ΔH is -899 J, P is 40 Pa, and ΔV is 1.0 m^3, we can substitute these values into the equation to calculate ΔU:
ΔU = -899 J - (40 Pa) * (1.0 m^3)
Note that the units of pressure must be converted to Joules (J) since the units of ΔH are also in Joules. To do this, we'll use the conversion:
1 Pascal (Pa) = 1 Joule/m^3 (J/m^3)
ΔU = -899 J - (40 J/m^3 * 1.0 m^3)
ΔU = -899 J - 40 J
ΔU = -939 J
Therefore, the internal energy for the reaction is -939 J.