One mole of ethanol $(l)$ was completely burnt in oxygen to form $CO_{2(g)}$ and $H_2O_{(l)}$. What is the $\Delta_r H^{\ominus}$ (in $kJ \ mol^{-1}$) for this reaction?
(The standard enthalpy of formation $(\Delta_f H^{\ominus})$ of $C_2H_5OH_{(l)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ is respectively $-277, -393$ and $-286 \ kJ \ mol^{-1}$.)
(The standard enthalpy of formation $(\Delta_f H^{\ominus})$ of $C_2H_5OH_{(l)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ is respectively $-277, -393$ and $-286 \ kJ \ mol^{-1}$.)
- A$+1921$
- B$-1921$
- C$+1367$
- D$-1367$
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Similar Questions
If $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$,then the heat of formation of $CO$ is
Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?
Medium
View SolutionCalculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.
Medium
View SolutionGiven
$(i) \, 2Fe_2O_{3(s)} \to 4Fe_{(s)} + 3O_{2(g)}$
$\Delta _rG^o = + 1487.0 \, kJ \, mol^{-1}$
$(ii) \, 2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$
$\Delta _rG^o = - 514.4 \, kJ \, mol^{-1}$
Free energy change,$\Delta _rG^o$ for the reaction
$2Fe_2O_{3(s)} + 6CO_{(g)} \to 4Fe_{(s)} + 6CO_{2(g)}$ will be ..... $kJ \, mol^{-1}$
$(i) \, 2Fe_2O_{3(s)} \to 4Fe_{(s)} + 3O_{2(g)}$
$\Delta _rG^o = + 1487.0 \, kJ \, mol^{-1}$
$(ii) \, 2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$
$\Delta _rG^o = - 514.4 \, kJ \, mol^{-1}$
Free energy change,$\Delta _rG^o$ for the reaction
$2Fe_2O_{3(s)} + 6CO_{(g)} \to 4Fe_{(s)} + 6CO_{2(g)}$ will be ..... $kJ \, mol^{-1}$
DifficultJEE MAIN 2018
View SolutionThe standard enthalpies of formation of $Al_{2}O_{3}$ and $CaO$ are $-1675 \ kJ \ mol^{-1}$ and $-635 \ kJ \ mol^{-1}$ respectively.
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).
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