5 grams of C2H5OH is burned in a bomb calorimeter, producing dqv = -148.3 kJ, by the
following reaction: C2H5OH(l) + 3O2(g) 6 2CO2(g) + 3H2O(l)
a) Determine ΔU for the reaction
b) Determine ΔH for the reaction.
a) ΔU for the reaction can be determined using the equation ΔU = q + w, where q is the heat transferred and w is the work done. In a bomb calorimeter, no work is done, so w = 0. Therefore, ΔU = q. Since q = -148.3 kJ, the ΔU for the reaction is also -148.3 kJ.
b) Since the reaction is taking place at constant volume in a bomb calorimeter, ΔU is equal to ΔH. So, ΔH for the reaction is also -148.3 kJ. However, be careful not to mix up ΔH with ΔH°, which is the change in enthalpy under standard conditions.
To solve this problem, we need to use the relationship between ΔU (internal energy change) and ΔH (enthalpy change). The equation is as follows:
ΔH = ΔU + PΔV
Where:
ΔH is the enthalpy change
ΔU is the internal energy change
P is the pressure
ΔV is the change in volume
In this case, we are using a bomb calorimeter. Bomb calorimeters are at constant volume, so there is no change in volume (ΔV = 0). Therefore, the equation simplifies to:
ΔH = ΔU
a) Determining ΔU for the reaction:
ΔU = -148.3 kJ (given)
b) Determining ΔH for the reaction:
ΔH = ΔU = -148.3 kJ
So, the ΔH for the reaction is -148.3 kJ.
To determine ΔU (change in internal energy) for the reaction, we need to use the first law of thermodynamics, which states that ΔU is equal to q (heat transfer) minus w (work done). In a bomb calorimeter, the system is at constant volume, so there is no work done (w = 0). Therefore, ΔU is equal to q.
Given that the heat transfer (q) is -148.3 kJ, we can conclude that ΔU is also -148.3 kJ.
To determine ΔH (change in enthalpy) for the reaction, we need to use the equation ΔH = ΔU + PΔV, where P is the pressure and ΔV is the change in volume. In this case, since the reaction is carried out in a bomb calorimeter, the pressure is constant, so PΔV is also equal to zero.
Therefore, ΔH for the reaction is equal to ΔU, which in this case is also -148.3 kJ.