Calculate the change in enthalpy, ΔH in joules (J) for the reaction below carried out at constant pressure of 11.1 atm at a volume of -2.3 m3 with internal energy of -90.9 kJ.
N2 +3H2 → 2NH3 ΔH = ?
-64.4 x 103 J
-116.4 x 103 J
+116.43 J
+ 64.4 J
The calculation involves using the equation:
ΔH = ΔE + PΔV
where ΔE is the change in internal energy, P is the constant pressure, and ΔV is the change in volume.
First, we need to convert the internal energy from kilojoules to joules:
ΔE = -90.9 kJ = -90.9 x 10^3 J
Next, we need to calculate the value of ΔV. The given volume of -2.3 m^3 suggests that there has been a decrease in volume, which means that ΔV is negative:
ΔV = -2.3 m^3
Now we can substitute these values into the equation and solve for ΔH:
ΔH = ΔE + PΔV
ΔH = -90.9 x 10^3 J + (11.1 atm)(-2.3 m^3)
ΔH = -90.9 x 10^3 J - 25.53 x 10^3 J
ΔH = -116.43 x 10^3 J
Therefore, the change in enthalpy for the given reaction at constant pressure is -116.4 x 10^3 J. The answer is B.
To calculate the change in enthalpy (ΔH) for the given reaction, you need to use the equation:
ΔH = ΔU + PΔV
Where:
ΔH = change in enthalpy
ΔU = change in internal energy
P = pressure
ΔV = change in volume
Given data:
ΔU = -90.9 kJ = -90.9 x 10^3 J (converted to joules)
P = 11.1 atm
ΔV = -2.3 m^3
First, let's convert the pressure from atm to Pa (Pascal):
1 atm = 101325 Pa
11.1 atm = 11.1 x 101325 Pa = 1.124275 x 10^6 Pa
Now, let's calculate the change in enthalpy using the equation:
ΔH = ΔU + PΔV
ΔH = -90.9 x 10^3 J + (1.124275 x 10^6 Pa x -2.3 m^3)
Calculating:
ΔH = -90.9 x 10^3 J + (-2.584℗251 x 10^6 J)
ΔH ≈ -90.9 x 10^3 J - 2.584 x 10^6 J
ΔH ≈ -2.6749 x 10^6 J
So the change in enthalpy, ΔH, is approximately -2.6749 x 10^6 J.