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Enthalpy Change MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

5 Questions around this concept.

Solve by difficulty

EASY

Consider the following reactions:

(i) $H^{+}_{(aq)}+OH^{-}_{(aq)}= H_2O_{(l)},$

$\Delta H=-X_1\, \, \, kJmol^{-1}$

(ii) $H_2_(g_) +\frac{1}{2}O_2_(g_)= H_2O_(l_)$

$\Delta H= -X_2\, \, kJmol^{-1}$

(iii) $CO_2_(g_) +H_2_(g_)= CO_(g_) +H_2O,$

$\Delta H=-X_3\: \: kJmol^{-1}$

(iv) $C_2H_2_(g_) +\frac{5}{2}O_2_(g_)= 2CO_2_(g_)+ H_2O_(l_)$

$\Delta H= +X_4\: \: kJmol^{-1}$

Enthalpy of formation of $H_2O_(l_)$ is

$+X_3\: \: kJmol^{-1}$

$-X_4\: \: kJmol^{-1}$

$+X_1\: \: kJmol^{-1}$

$-X_2\: \: kJmol^{-1}$

View Solution

Given C_(graphite)+O₂(g) → CO₂(g) ;

_rH⁰=−393.5 kJ mol⁻¹

H₂(g)+ $\frac{1}{2}$ O₂(g) → H₂O(l) ;

rH⁰=−285.8 kJ mol⁻¹

CO₂(g)+2H₂O(l) → CH₄(g)+2O₂(g) ;

_rH⁰=+890.3 kJ mol⁻¹

Based on the above thermochemical equations, the value of _rH⁰ at 298 K for the reaction

C_(graphite)+2H₂(g) → CH₄(g) will be :

−74.8 kJ mol⁻¹

−144.0 kJ mol⁻¹

+74.8 kJ mol⁻¹

+144.0 kJ mol⁻¹

View Solution

Concepts Covered - 1

Standard Enthalpy And Enthalpy Of Formation

Heat of Formation

The amount of heat evolved or absorbed or change in enthalpy when 1 mole of a substance is obtained from its constituents or free elements.

$\Delta \mathrm{H}^{\circ}=\sum \mathrm{H}^{\circ} \mathrm{p}-\sum \mathrm{H}^{\circ} \mathrm{r}$

Once the value of H^o at 25^o C for any species has been assigned the value of H^o at other temperatures Can be found out by using Kirchoff's equation as follows.

$\int_{298}^{\mathrm{T}} \mathrm{dH}^{\circ}=\int_{298}^{\mathrm{T}} \mathrm{CpdT}$

$\mathrm{H_T^\circ}-\mathrm{H_{298}^\circ}=\int_{298}^{\mathrm{T}} \mathrm{Cp}\mathrm{d} \mathrm{T}$

1. Standard heat of formation of a free element is taken as zero.

For example, In carbon—graphite form is taken as standard state and in sulphur, the monoclinic form is standard state.

2. The heat of formation may be +ve or —ve.

$\begin{array}{l}{\text { If } \Delta \mathrm{H} \text { is -ve compound is exothermic. }} \\ {\text { If } \Delta \mathrm{H} \text { is +ve compound is endothermic. }}\end{array}$

3. The stability of exothermic compound is more than that of the endothermic compound hence, greater the liberated energy greater is the stability of the compound.

Example, HF > HCI > HBr > HI