For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$,$\Delta H^o = -198 \, kJ$. According to Le Chatelier's principle,the favorable conditions for the forward reaction are:

Consider the following reactions. In which cases is product formation favoured by decreased temperature?
$(a) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}; \ \Delta H^o = 181 \ kJ$
$(b) \ 2CO_{2(g)} \rightleftharpoons 2CO_{(g)} + O_{2(g)}; \ \Delta H^o = 566 \ kJ$
$(c) \ H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}; \ \Delta H^o = -9.4 \ kJ$
$(d) \ H_{2(g)} + F_{2(g)} \rightleftharpoons 2HF_{(g)}; \ \Delta H^o = -541 \ kJ$

Explain the effect of adding $(i)$ Oxalic acid $(H_2C_2O_4)$,$(ii)$ $HgCl_2$,and $(iii)$ Potassium thiocyanate $(KSCN)$ on the equilibrium reaction: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$ (deep red color).

$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction:
$3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4OH^-$
The reaction can go to completion by removing $OH^-$ ions by adding:

If pressure increases,what is the effect on the given equilibrium $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$?

In the reaction ${A_2}_{(g)} + 4{B_2}_{(g)} \rightleftharpoons 2A{B_4}_{(g)}$ where $\Delta H < 0$,the formation of $AB_{4(g)}$ will be favoured by: