Question

2a + b → 3c + d δh = 313 kj
determine the enthalpy change, δh, for each reaction.
3c + d → 2a + b
δh = kj
a + \frac{1}{2}b → \frac{3}{2}c + \frac{1}{2}d
δh = kj
9c + 3d → 6a + 3b
δh = kj

Explanation:

Step1: Reverse the reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
For \(3C + D
ightarrow2A + B\), since \(2A + B
ightarrow3C + D\) has \(\Delta H = 313\space kJ\), then \(\Delta H=- 313\space kJ\)

Step2: Multiply the reaction by a fraction

When a reaction is multiplied by a fraction \(n\), \(\Delta H\) is also multiplied by \(n\).
For \(A+\frac{1}{2}B
ightarrow\frac{3}{2}C+\frac{1}{2}D\), the original reaction \(2A + B
ightarrow3C + D\) (\(\Delta H = 313\space kJ\)) is multiplied by \(\frac{1}{2}\). So \(\Delta H=\frac{313}{2}=156.5\space kJ\)

Step3: Multiply the reaction by an integer

When a reaction is multiplied by an integer \(n\), \(\Delta H\) is also multiplied by \(n\).
For \(9C + 3D
ightarrow6A + 3B\), the reversed reaction \(3C + D
ightarrow2A + B\) (\(\Delta H=-313\space kJ\)) is multiplied by \(3\). So \(\Delta H=-313\times3=-939\space kJ\)

Answer:

\(-313\space kJ\)
\(156.5\space kJ\)
\(-939\space kJ\)