A chemical reaction is run in which 736 Joules of work is done by the system and the internal energy changes by -1370 Joules.
pls help me
I’m guessing you need the Rxn Enthalpy. It’s the only given data item missing for the 1st Law equation…
ΔU = q + w => q = ΔU – w = (-1370 J) – (- 736 J) = (-1370 + 736) J = - 634 Joules exothermic heat of rxn.
ΔU = -1370 Joules (given)
w = -736 Joules (expansion work)*
*expansion => system is working (pushing back) on the surroundings => w < 0.
*compression => surroundings working (compressing) on the system => w > 0.
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Sure, I can help you with that! To solve this problem, we need to use 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:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
In this case, we are given that the work done by the system (W) is 736 Joules and the change in internal energy (ΔU) is -1370 Joules. We need to find the heat added to the system (Q).
Plugging the given values into the equation, we have:
-1370 J = Q - 736 J
To solve for Q, we can rearrange the equation:
Q = -1370 J + 736 J
Now we can calculate the value of Q:
Q = -634 Joules
Therefore, the heat added to the system is -634 Joules. The negative sign indicates that the system has lost heat to the surroundings.