Assume that a particular reaction evolves 230 kJ of heat and that 28 kJ of PV work is gained by the system.
What are the values of delta E and delta H for the system?
For the surroundings?
To determine the values of delta E and delta H for the system, we need to use the following formula:
delta E = q + w
where delta E is the change in internal energy of the system, q is the heat transferred to or from the system, and w is the work done on or by the system.
Given that the reaction evolves 230 kJ of heat (q = -230 kJ) and gains 28 kJ of PV work (w = 28 kJ), we can calculate the change in internal energy for the system:
delta E = (-230 kJ) + (28 kJ) = -202 kJ
Therefore, the value of delta E for the system is -202 kJ.
To determine the value of delta H for the system, we use the equation:
delta H = delta E + PdeltaV
where delta H is the change in enthalpy of the system, delta E is the change in internal energy, P is the pressure, and delta V is the change in volume.
Since we are not given specific values for pressure or volume, we cannot calculate delta H accurately with the given information.
Now, let's consider the surroundings. For the surroundings, the heat flow is opposite to that of the system. So, the value of q for the surroundings is +230 kJ, as we know the reaction evolves 230 kJ of heat.
For the surroundings, since they are not involved in any work, the value of w is zero.
Therefore, the change in internal energy for the surroundings (delta E_surroundings) is:
delta E_surroundings = q + w
= (+230 kJ) + (0 kJ)
= +230 kJ
Similarly, since the surroundings are not involved in any volume change, the value of delta V for the surroundings is zero.
Hence, the value of delta H for the surroundings (delta H_surroundings) is:
delta H_surroundings = delta E_surroundings + PdeltaV_surroundings
= (+230 kJ) + (0 kJ)
= +230 kJ
Therefore, the values of delta E for the surroundings and delta H for the surroundings are +230 kJ and +230 kJ respectively.
To determine the values of ΔE (change in internal energy) and ΔH (change in enthalpy) for the system and surroundings, we need to remember the following equations:
ΔE = q + w
ΔH = ΔE + PΔV
where ΔE is the change in internal energy, ΔH is the change in enthalpy, q is the heat absorbed or released, w is the work done, P is the pressure, and ΔV is the change in volume.
Given:
Heat evolved by the system (q) = -230 kJ (negative sign indicates heat is released by the system)
PV work gained by the system (w) = +28 kJ (positive sign indicates work is done on the system, so the system gains work)
Now, let's calculate the values for the system and surroundings:
1. ΔE for the system:
ΔE = q + w
ΔE = -230 kJ + 28 kJ
ΔE = -202 kJ
Therefore, the change in internal energy of the system (ΔE) is -202 kJ.
2. ΔH for the system:
ΔH = ΔE + PΔV
In this case, since we are not given the pressure and volume change, we cannot determine the exact value of ΔH. However, if we assume that the reaction is carried out at constant pressure, then ΔH is approximately equal to ΔE.
Therefore, the change in enthalpy of the system (ΔH) is approximately -202 kJ.
3. ΔE for the surroundings:
Since the system releases heat (-230 kJ), the surroundings gain heat of the same magnitude, but with the opposite sign:
ΔE_surroundings = -q_system
ΔE_surroundings = -(-230 kJ)
ΔE_surroundings = +230 kJ
Therefore, the change in internal energy of the surroundings (ΔE_surroundings) is +230 kJ.
4. ΔH for the surroundings:
ΔH_surroundings = ΔE_surroundings + PΔV
Similarly to the system, without more information about the pressure and volume change, it is difficult to determine the exact value of ΔH for the surroundings. However, assuming constant pressure, ΔH_surroundings is approximately equal to ΔE_surroundings.
Therefore, the change in enthalpy of the surroundings (ΔH_surroundings) is approximately +230 kJ.
In summary:
- ΔE for the system = -202 kJ
- ΔH for the system ≈ -202 kJ
- ΔE for the surroundings = +230 kJ
- ΔH for the surroundings ≈ +230 kJ