NEET Practice Questions (MCQs) with Answers & Solutions

Heat energy is also involved when one allotropic form of an element is converted in to another. graphite is the stabler allotrope because the heat of transformation of $C_{\left(diamond\right)}\to C_{\left(graphite\right)}$.

(i) $C_{\left(dia\right)}+O_{2\left(g\right)}=C{O}_{2\left(g\right)}\Delta H=-94.5kcal$

(ii) $C_{\left(graphite\right)}+O_{2\left(g\right)}=C{O}_{2\left(g\right)}\Delta H=-94.0kcal$

$\Delta H_{transformation}=-94.5-(-94.0)$

= –0.5 k cal.

All are the correct statements.

Density is mass per unit volume and specific volume is volume per unit mass.

The heat of formation of HCl is to be calculated on per mole of HCl produced

$\therefore$ heat of formation of HCl = $-\frac{44}{2}$ = -22 kcal.

The statement is definition of Hess's law

$∆H=∆E+P∆V$

1.00 = $$\begin{gathered} ∆E+\frac{1\left(1.5-1.2\right)L atm}{0.082 L atm}\times 2\times {10}^{-3} kcal \\ ∆E=0.993 kcal \end{gathered}$$

The heat of combustion of per gm of $C{H}_4$

= $-\frac{890}{16}=-55.625 kJ$

The heat of combustion of per gm of $C_2{H}_4$

=$\frac{1411}{28}=-50.393 kJ$

The heat of combustion of per gm of $C_2{H}_6$

=$-\frac{1560}{30}=-52 kJ$

The standard heat of combustion is defined as the heat evolved on burning one mole of the compound in presence of O₂ to give CO₂ and water. Since the amount of the substance is specified, it becomes an intensive property.

Kirchhoff's equation is:

$∆H_2-∆H_1=∆C_P\left(T_2-T_1\right)$

When 5 ml of strong acid reacting with 5 ml of strong base net, heat evolved = $∆H$

$\therefore$ temperature increase = $\frac{∆H}{ms}$

when 10 ml of each are reaction, the heat evolved = $2∆H;$

but mass of the solution = 2m

$\therefore$ temperature increase = $\frac{2∆H}{2ms}=\frac{∆H}{ms}$