The heats of hydrogenation for $3-$-methylbutene and $2-$-pentene are $-30\, kcal/mol$ and $-28\, kcal/mol$ respectively. The heats of combustion of $2-$-methylbutane and pentane are $-784\, kcal/mol$ and $-782\, kcal/mol$ respectively. All the values are given under standard conditions. Taking into account that combustion of both alkanes gives the same products,what is $\Delta H$ (in $kcal/mol$) for the following reaction under same conditions?
$3-$-methylbutene $\rightleftharpoons$ $2-$-pentene
$3-$-methylbutene $\rightleftharpoons$ $2-$-pentene
- A$0$
- B$-4$
- C$-2$
- D$2$
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Similar Questions
Calculate the heat required to convert $9 \ g$ of liquid water to water vapor using the following equations:
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)} \quad \Delta H = -57 \ kCal$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H = -68.3 \ kCal$ (in $kCal$)
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)} \quad \Delta H = -57 \ kCal$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H = -68.3 \ kCal$ (in $kCal$)
Based on the values of $B.E.$ given,calculate the $\Delta_fH^o$ of $N_2H_4\, (g)$ in $kJ\, mol^{-1}$.
Given : $N-N = 159\, kJ\, mol^{-1};\, H-H = 436\, kJ\, mol^{-1}$
$N \equiv N = 941\, kJ\, mol^{-1};\, N-H = 398\, kJ\, mol^{-1}$
Given : $N-N = 159\, kJ\, mol^{-1};\, H-H = 436\, kJ\, mol^{-1}$
$N \equiv N = 941\, kJ\, mol^{-1};\, N-H = 398\, kJ\, mol^{-1}$
Medium
View SolutionCalculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
DifficultTS EAMCET 2007
View SolutionCalculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$
If the enthalpies of formation of $Al_2O_3$ and $Cr_2O_3$ are $-1596 \, kJ/mol$ and $-1134 \, kJ/mol$ respectively,then calculate $\Delta H$ for the following reaction in $kJ$:
$2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$
$2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$
Medium
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