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	${\Delta }_{\text{vap }}H=88J{K}^{-1}mo{l}^{-1}\times$ 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)

$(a\to s)$ Hess' law states that enthalpy change in a reaction remains the same whether the reaction takes place in one step or in several steps
$(b\to r)$ Combustion reactions are exothermic
$(c\to q)\frac{{\Delta }_{\text{vap H}}}{\text{ Boiling point in }K}=88J{K}^{-1}mo{l}^{-1}$
$(d\to p)2.303\log \frac{P_2}{P_2}=\frac{{\Delta }_{vap}H}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
It is Clausius-Clapeyron equation