Question

For reaction 2A(g) + B(g) $⟶$ 3C (g) $\Delta$H = $-x$ kJ

then for reaction $\frac{3}{2}$ C(g) $⟶$ A(g) + $\frac{B}{2}$ (g) $\Delta$H = $\frac{x}{2}$ kJ.

• A. True
• B. False

Answer 1

The correct option is A True

The given statement is true.
When the reaction is reversed, the enthalpy change of the reaction changes its sign from minus to plus. Hence, $\Delta H=+x$ kJ
When the number of moles of each reactant and product are reduced to one half, the magnitude of the enthalpy change for the reaction is also reduced to one half.
Hence, $\Delta H=\frac{+x}{2}$ kJ