Change in the enthalpy of a reaction at constant pressure is given by

Options

A. `DeltaE`
B. `DeltaH`
C. `PDeltaV`
D. `DeltaC_(P)`

Answer

To find the change in enthalpy of a reaction at constant pressure, we start with the definition of enthalpy (H):

1. **Definition of Enthalpy**:
\[ H = U + PV \]
where \( H \) is enthalpy, \( U \) is internal energy, \( P \) is pressure, and \( V \) is volume.

2. **Differentiating Enthalpy**:
When we consider a small change in enthalpy, we can express it as:
\[ dH = dU + d(PV) \]
Expanding \( d(PV) \) using the product rule gives:
\[ d(PV) = PdV + VdP \]
Therefore, we can rewrite the change in enthalpy as:
\[ dH = dU + PdV + VdP \]

3. **Applying the First Law of Thermodynamics**:
According to the first law of thermodynamics:
\[ dU = dQ - dW \]
where \( dQ \) is the heat added to the system and \( dW \) is the work done by the system. For expansion work at constant pressure, we have:
\[ dW = PdV \]
Substituting this into the expression for \( dU \):
\[ dU = dQ - PdV \]

4. **Substituting Back into the Enthalpy Equation**:
Now, substituting \( dU \) back into the equation for \( dH \):
\[ dH = (dQ - PdV) + PdV + VdP \]
The \( PdV \) terms cancel out:
\[ dH = dQ + VdP \]

5. **Considering Constant Pressure**:
At constant pressure, the change in pressure \( dP \) is zero, so the term \( VdP \) becomes zero:
\[ dH = dQ \]

6. **Conclusion**:
Therefore, the change in enthalpy at constant pressure is equal to the heat added to the system:
\[ \Delta H = Q \text{ (at constant pressure)} \]

Thus, the change in enthalpy of a reaction at constant pressure is given by \( \Delta H = Q \).