Question

$\begin{array}{cccccc}& Columns1represents\Delta G,& & Column2\Delta H& & andColumn3\Delta S.\\ & Column1& & Column2& & Column3\\ \left(I\right)& +ve& \left(i\right)& -ve& \left(P\right)& \infty \\ \left(II\right)& \infty & \left(ii\right)& 0& \left(Q\right)& +ve\\ \left(III\right)& 0& \left(iii\right)& \infty & \left(R\right)& -ve\\ \left(IV\right)& -ve& \left(iv\right)& +ve& \left(S\right)& 0\end{array}$
The correct set for Haber’s process is (when equilibrium is just to be reached)

• A. (IV) (i) (R)
• B. (IV) (i) (P)
• C. (II) (iii) (R)
• D. (III) (ii) (Q)

Answer 1

The correct option is A (IV) (i) (R)

The Haber’s process is given as:
$\underset{\left(g\right)}{N_2}+\underset{\left(g\right)}{3H_2}\underset{Mo\left(s\right)}{Fe\left(s\right)}2\underset{\left(g\right)}{N{H}_3}+Heat$
Before reaching to the equilibrium, this reaction is favourable in forward direction
$\Rightarrow \Delta G=-ve.$
The reaction suggests that heat is released $\Rightarrow \Delta H=-ve.$
In the reactant side the no. of gaseous moles are higher and lesser in the product side.
It means entropy is decreased.
$\Rightarrow \Delta S=-ve$