Question

When 2 moles of C2H6 are completely burnt 3129 KJ of heat is liberated. Calculate the heat of formation, ΔHf for C2H6; ΔHf for CO2­ and H2O are -395 and -286 KJ respectively.

Answer 1

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Answer 2

The heat of formation of $C_2H_6$ is calculated as -83.5KJ

Given:

Heat liberated (ΔH°) = 3129 KJ

ΔH of $CO_2$ = -395 KJ

ΔH of $H_2O$ = -286 KJ

To find:

ΔH of $C_2H_6$ = ?

Formula to be used:

$\Delta H^{0}=\Delta H_{f(\text { Product })}-\Delta H_{f(\text { Reactants })}$

Chemical equation:

• The balanced chemical reaction involved in combustion of 2 moles of $C_2H_6$ is as follows:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

• From the above-given chemical equation it can be inferred that for combustion of 2 moles of $C_2H_6$, 7 moles of $O_2$ is required.
• The combustion yileds 4 moles of $CO_2$ and 6 moles of $H_2O$.
• In the reaction:

Reactants : $C_2H_6$ and $O_2$

Products : $CO_2$ and $H_2O$

Calculation:

$\Delta H^{0}=\Delta H_{f(\text { Product })}-\Delta H_{f(\text { Reactants })}$

$3129 = [4\times\Delta H_{(CO_2)} + 6\times\Delta H_{(H_2O)}]-[2\times\Delta H_{(C_2H_6)}+7\times\Delta H_{(O_2)}]$

$3129 = [4\times(-395)+ 6\times(-286)]-[2\times\Delta H_{(C_2H_6)}+7\times 0]$

$[2\times\Delta H_{(C_2H_6)}+7\times 0] = [4\times(-395)+ 6\times(-286)] - 3129$

$[2\times\Delta H_{(C_2H_6)}] = [(-1580)+ (-1716)] - 3129$

$[2\times\Delta H_{(C_2H_6)}] = 167$

$\Delta H_{(C_2H_6)} = \frac{167}{2}$

$\Delta H_{(C_2H_6)} = -83.5$

Conclusion:

Thus the heat of formation of $C_2H_6$ is calculated as -83.5KJ.

Learn more about such concept

2. Heat of formation of benzene, assuming no

resonance. Given that

BE (C-C) = 83 kcal

BE (C=C) = 140 kcal

BE (C-H) = 99 kcal

Heat of atomisation of C = 170.9 kcal

Heat of atomisation of H = 104.2 kcal

will be

(1) 39 kcal

(2) 75 kcal

(3) 1263 kcal

(4) 421 kcal​

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Define standard enthalpy of formation . write a chemical equation for the formation of N2O5. answer in 100 to 150 words.

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