What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
(N/A) Exothermic reactions: Reactions which evolve heat are known as exothermic reactions. Evolution of heat can be shown using a '$+$' sign or by a $\Delta H$ with a '$-$' sign along with the product.
Example: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} + 393.5 \ kJ$
Endothermic reactions: Reactions which absorb heat are called endothermic reactions. Absorbed heat can be shown as a '$-$' sign or in terms of $\Delta H$ with a '$+$' sign.
Example: $C_{(s)} + H_{2}O_{(g)} \longrightarrow CO_{(g)} + H_{2(g)} - 131.4 \ kJ$
Example: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} + 393.5 \ kJ$
Endothermic reactions: Reactions which absorb heat are called endothermic reactions. Absorbed heat can be shown as a '$-$' sign or in terms of $\Delta H$ with a '$+$' sign.
Example: $C_{(s)} + H_{2}O_{(g)} \longrightarrow CO_{(g)} + H_{2(g)} - 131.4 \ kJ$
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Similar Questions
For the reaction $2 H_2 + O_2 \rightarrow 2 H_2 O$,$\Delta H = -571 \ kJ$. Bond energy of $H-H = 435 \ kJ$ and $O=O = 498 \ kJ$. Then the average bond energy of $O-H$ bond will be:
Based on the following thermochemical equations,find the value of $x$ in $kJ$.
$(i) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} ; \Delta H = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} ; \Delta H = -242 \ kJ$
$(iv) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -x \ kJ$
$(i) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} ; \Delta H = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} ; \Delta H = -242 \ kJ$
$(iv) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -x \ kJ$
Difficult
View SolutionThe enthalpy change at $298 \ K$ for the decomposition of water is given in the following two steps:
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$
Medium
View SolutionCalculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
DifficultTS EAMCET 2008
View SolutionIf the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$
Difficult
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