When 1 mol of a gas burns at constant pressure, it produces 2426 J of heat and does 6 kJ of work.
What is Delta H, q, and w?
H=?
q=2.426 kJ
w=-6 kJ
Therefore, E=-8
To find the values of ΔH (change in enthalpy), q (heat), and w (work), we can use the First Law of Thermodynamics, which states that the internal energy change of a system is equal to the heat added to the system minus the work done by the system.
The equation is as follows:
ΔU = q - w
Here, ΔU represents the change in internal energy. For a system at constant pressure, ΔU is equal to ΔH (change in enthalpy). So, we can make the following substitution:
ΔH = q - w
Given that 1 mol of the gas burns at constant pressure, producing 2426 J of heat (q) and doing 6 kJ of work (w), we can substitute these values into the equation:
ΔH = 2426 J - 6000 J
Converting 6 kJ to J, we multiply 6 kJ by 1000 to get 6000 J.
Now we can calculate ΔH:
ΔH = -3574 J
Therefore, ΔH is equal to -3574 J, q is equal to 2426 J, and w is equal to 6000 J. The negative sign indicates that the process is exothermic since heat is being released.