Question

given the thermochemical equations
a(g)→b(g) δh=80kj
b(g)→c(g) δh=−120kj
find the enthalpy changes for each reaction.
3a(g)→3b(g) δh=

b(g)→a(g) δh=

a(g)→c(g) δh=

Explanation:

Step1: For \(3A(g)\to3B(g)\)

According to Hess's law, if we multiply a reaction by a factor \(n\), the enthalpy change \(\Delta H\) is also multiplied by \(n\).
Given \(A(g)\to B(g)\), \(\Delta H = 80\space kJ\).
For \(3A(g)\to3B(g)\), \(n = 3\).
So \(\Delta H=3\times80\space kJ\)

Step2: For \(B(g)\to A(g)\)

If we reverse a reaction, the sign of \(\Delta H\) changes.
Given \(A(g)\to B(g)\), \(\Delta H = 80\space kJ\).
For \(B(g)\to A(g)\), \(\Delta H=- 80\space kJ\)

Step3: For \(A(g)\to C(g)\)

Using Hess's law, if \(A(g)\to B(g)\), \(\Delta H_1 = 80\space kJ\) and \(B(g)\to C(g)\), \(\Delta H_2=-120\space kJ\)
Then \(A(g)\to C(g)\) is the sum of \(A(g)\to B(g)\) and \(B(g)\to C(g)\)
\(\Delta H=\Delta H_1+\Delta H_2=80+( - 120)\space kJ\)

Answer:

For \(3A(g)\to3B(g)\), \(\Delta H = 240\space kJ\)
For \(B(g)\to A(g)\), \(\Delta H=-80\space kJ\)
For \(A(g)\to C(g)\), \(\Delta H=-40\space kJ\)