Question

Column I		Column II
A	Hess' law	p	$2.303 \log \frac{P_2}{P_1}=\frac{\Delta_{\text {vap }} H}{R}\left(\frac{T_2-T_1}{T_1 T_2}\right)$
B	Combustion reaction	q	$\Delta_{\text {vap }} H=88 \,J \,K ^{-1}\, mol ^{-1} \times$ Boiling point in Kelvin
C	Trouton's law	r	Exothermic
D	Clausius-Cal-peyron equation	s	$\Delta H$ remains the same irrespective steps

• A. A= s ,B= r ,C= q ,D= p
• B. A= r ,B=s ,C=p ,D= q
• C. A=p ,B= q ,C= s,D= r
• D. A= q ,B=p ,C=r ,D= s

Answer 1

Answer: (A)

$( a \rightarrow s )$ Hess' law states that enthalpy change in a reaction remains the same whether the reaction takes place in one step or in several steps
$( b \rightarrow r )$ Combustion reactions are exothermic
$( c \rightarrow q ) \frac{\Delta_{ \text{vap H}}}{\text { Boiling point in } K }=88\, J K ^{-1}\, mol ^{-1}$
$( d \rightarrow p ) 2.303 \log \frac{P_2}{P_2}=\frac{\Delta_{ vap } H}{R}\left(\frac{T_2-T_1}{T_1 T_2}\right)$
It is Clausius-Clapeyron equation