(i) Derive the relation between $K_p$ and $K_c$ for general homogeneous gaseous reaction.
(ii) How do you measure heat changes of a constant pressure ?

Consider a gerneral reaction in which all reactants and products are ideal gases.
$xA+yB\iff lC+mD$
The equilibrium costant $K_c$ is
$K_c=\frac{{\left[C\right]}^l\left[D\right]}{{\left[A\right]}^x{\left[B\right]}^y}$
$K_p=\frac{p^{l}C\times p_D^m}{p_A^x\times p_B^y}$
The ideal gas equation is
$PV=nRT\text{or}P=\frac{n}{V}RT$
Since, Active mass=molar concentration $=\frac{m}{v}$
P= Active mass $\times$ RT
Based on the above expressio, the partial pressure of the reacants and products can be expressred as,
$p_A^x={\left[A\right]}^x.{\left[RT\right]}^x,p_B^y={\left[B\right]}^y.{\left[RT\right]}^y$
$p_C^l={\left[C\right]}^l.{\left[RT\right]}^l,P_D^m={\left[D\right]}^m.{\left[RT\right]}^m$
On substiutiono in equation (2),
$K_p=\frac{{\left[C\right]}^l{\left[RT\right]}^l{\left[D\right]}^m{\left[RT\right]}^m}{{\left[A\right]}^x{\left[RT\right]}^x{\left[B\right]}^y{\left[RT\right]}^y}$
$K_p=\frac{{\left[C\right]}^l{\left[D\right]}^m}{{\left[A\right]}^x{\left[B\right]}^y}\times R{T}^{(l+m)-(x+y)}$
By comparing equation (1) and (4) ,we get
$K_p=K_c{\left(RT\right)}^{\Delta ng}$
where $\Delta n_g$ is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If $\Delta n_g=0,K_p=K_c{\left(RT\right)}^0$
$K_p=K_c$
Example : $H_2\left(g\right)+I_2\left(g\right)\iff 2HI\left(g\right)$
(ii) when $\Delta n_g=+ve$
$K_p=K_c{\left(RT\right)}^{+ve}\Rightarrow K_p=K_c$
Example $2N{H}_3\left(g\right)\iff N_2\left(g\right)+3H_2\left(g\right)$
(iii) When $\Delta n_g=-ve$
$K_p=K_c{\left(RT\right)}^{-ve}$
$K_p<K_c$
Example : $2S{O}_2\left(g\right)+O_2\left(g\right)\iff 2S{O}_3\left(g\right)$
1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2 . We know that $\Delta H=q_p$ (at constant P) and therefore , heat absorbed or evoloved $q_r$ constant pressure is also called tha heat of reaction or enthalpy of reaction $\Delta H_r$
3. In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. Therefore $q_p$ will be negative and $\Delta H_r$ will also be negative
4. Similarly in an endothermic reaction, heat is absorbed $q_p$ is postive and $\Delta H_r$ will also be positive.