If a particular reaction absorbs 252 kJ of heat while doing 105 kJ of work at constant pressure. What are the ΔE and ΔH for values for the reaction? Is the reaction endothermic or exothermic?
q = 252 kJ and it's + so is endothermic
w = 105 kJ and it's - since it is doing the work.
delta E = q+w
delta E = +252 kJ + (-105 kJ)
what about the delta H?
delta H at constant pressure is q. Or perhaps I should say it another way,k q at constant pressure is delta H. So delta H in this case is +252 kJ.
To find the values of ΔE (change in internal energy) and ΔH (change in enthalpy) for the reaction, we need to apply the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔE = q - w
ΔH, on the other hand, represents the change in enthalpy, which is equal to the heat absorbed or released during a reaction occurring at constant pressure. In other words, ΔH = q, assuming no non-PV work is done.
Given that the reaction absorbs 252 kJ of heat (q = 252 kJ) and does 105 kJ of work (w = -105 kJ, as work done by the system is negative), we can calculate the values for ΔE and ΔH:
ΔE = q - w
= 252 kJ - (-105 kJ)
= 252 kJ + 105 kJ
= 357 kJ
ΔH = q
= 252 kJ
Therefore, the values for the reaction are ΔE = 357 kJ and ΔH = 252 kJ.
To determine whether the reaction is endothermic or exothermic, we examine the sign of ΔH. If ΔH is positive, the reaction is endothermic, meaning it absorbs heat from the surroundings. Conversely, if ΔH is negative, the reaction is exothermic, meaning it releases heat to the surroundings.
In this case, since ΔH = 252 kJ (which is positive), the reaction is endothermic.