When 1mol of a gas burns at constant pressure, it produces 2430J of heat and does 5kJ of work.
Calculate change in H, q, and in kJ?
Re-read your post. Did you omit calculating dE?
2430 J and it is exothermic. q = -2430J so dH is -2430 J.
It does 5,000 J work so work - 5000 J.
dE = q + w. dE will be in J.
Convert to kJ and convert the other so to kJ.
Thank you so much! on dE i get 2.57 but it is telling me it is wrong?
never mind I got it (:
Both are negative. Why did you subtract?
dE = -2.43-5.00 = -7.43 kJ
To calculate the change in enthalpy (ΔH), heat (q), and work (w), we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system minus the work done by the system:
ΔU = q - w
Since we are given the values of heat (2430 J) and work (5 kJ = 5000 J), we can substitute these values into the equation to solve for ΔU:
ΔU = 2430 J - 5000 J
ΔU = -2570 J
The negative sign indicates that the system has lost internal energy. Since the change in enthalpy (ΔH) is equal to the change in internal energy at constant pressure, we can say:
ΔH = ΔU = -2570 J
To convert the energy into kilojoules (kJ), we divide the value by 1000:
ΔH = -2570 J / 1000
ΔH = -2.57 kJ
So, the change in enthalpy is -2.57 kJ, indicating a decrease in enthalpy.
To summarize:
- Change in enthalpy (ΔH) = -2.57 kJ
- Heat (q) = 2430 J
- Work (w) = 5000 J