When 2 moles of C2H6 are completely burnt- 3120 kJ of heat is liberated- Calculate the heat of formation -Hof for C2H6-Given- -Hof for CO2g and H2Ol are -395 and -286kJ respectively-

The correct option is A 88 kJ/mol 2C_2H_6+O_2⟶ 2CO_2+3H_2O Δ H_rxn=Δ Hmole = 31202= 1560 kJ Δ H_rxn=2( 345)+3( 286) (ΔH_f^o)_C_2H_6 Δ ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy of Combustion

When 2 mole...

Question

When $2$ moles of $C_{2} H_{6}$ are completely burnt, $3120 k J$ of heat is liberated. Calculate the heat of formation $Δ H_{f}^{o}$ for $C_{2} H_{6}$ .
Given: $Δ H_{f}^{o}$ for ${C O}_{2} (g)$ and $H_{2} O (l)$ are $- 395$ and $- 286 k J$ respectively.

- 88 k J / m o l

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+ 88 k J / m o l

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- 58 k J / m o l

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None of the above

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Solution

The correct option is A $- 88 k J / m o l$
$2 C_{2} H_{6} + O_{2} ⟶ 2 {C O}_{2} + 3 H_{2} O$

$Δ H_{r x n} = \frac{Δ H}{m o l e} = \frac{3120}{2} = - 1560 k J$

$Δ H_{r x n} = 2 (- 345) + 3 (- 286) - {(Δ H_{f}^{o})}_{C_{2} H_{6}}$

$Δ H_{f}^{o} = - 790 - 858 + 1560$ $= - 88 k J / m o l$

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Similar questions

Q. When

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C_{2} H_{6}

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3120 k J

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Q. When 2 moles of

C_{2} H_{6} (g)

are completely burnt 3120 kJ of heat is liberated. The enthalpy of
formation, of

C_{2} H_{6} (g)

in kJ / mol is X. Then -X is
Given

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C O_{2} (g)

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When 2 mole of $C_{2} H_{6}$ are completely burnt, 3129 KJ of heat is liberated. Calculate the heat of formation of $C_{2} H_{6} . △ H_{f}$ for $C O_{2} & H_{2} O$ are -395 KJ and -286 KJ respectively.

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When 2 moles of C2H6 are completely burnt, 3120 kJ of heat is liberated. Calculate the heat of formation ΔHof for C2H6.Given: ΔHof for CO2(g) and H2O(l) are −395 and −286kJ respectively.

The correct option is A −88 kJ/mol2C2H6+O2⟶2CO2+3H2OΔHrxn=ΔHmole=31202=−1560 kJΔHrxn=2(−345)+3(−286)−(ΔHof)C2H6ΔHof=−790−858+1560=−88kJ/mol

When $2$ moles of $C_{2} H_{6}$ are completely burnt, $3120 k J$ of heat is liberated. Calculate the heat of formation $Δ H_{f}^{o}$ for $C_{2} H_{6}$ .
Given: $Δ H_{f}^{o}$ for ${C O}_{2} (g)$ and $H_{2} O (l)$ are $- 395$ and $- 286 k J$ respectively.

The correct option is A $- 88 k J / m o l$
$2 C_{2} H_{6} + O_{2} ⟶ 2 {C O}_{2} + 3 H_{2} O$

$Δ H_{r x n} = \frac{Δ H}{m o l e} = \frac{3120}{2} = - 1560 k J$

$Δ H_{r x n} = 2 (- 345) + 3 (- 286) - {(Δ H_{f}^{o})}_{C_{2} H_{6}}$

$Δ H_{f}^{o} = - 790 - 858 + 1560$ $= - 88 k J / m o l$