Given the following- Cs - O2g - CO2g- - H - -394 kJ-mol 2H2g - O2g - 2H2Ol- - H - -568 kJ-mol C2H5OHl-3O2g- 2CO2g - 3H2Ol- - H - -1058 kJ -molUsing the given data- the heat of formation of ethanol is-

The correct option is C 582 kJ/mol heat of formation for ethnaol 2C + 3H_2 + 12O_2 ⟶ C_2H_5OH To get the enthalpy from the above reactions,Multiply ...

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Byju's Answer

Standard XII

Chemistry

Heat of Formation

Given the fol...

Question

Given the following:
$C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 394 k J / m o l$
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); Δ H = - 568 k J / m o l$ $C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l); Δ H = - 1058 k J / m o l$
Using the given data, the heat of formation of ethanol is:

- 582 k J / m o l

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71.8 k J / m o l

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- 244 k J / m o l

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+ 782 k J / m o l

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Solution

The correct option is C $- 582 k J / m o l$
heat of formation for ethnaol-
$2 C + 3 H_{2} + \frac{1}{2} O_{2} ⟶ C_{2} H_{5} O H$
To get the enthalpy from the above reactions,
Multiply the first reaction and second reaction with $2$ and $\frac{3}{2}$ respectively and reverse the third reaction after that add all the enthalpies-
$2 C + 2 O_{2} \to 2 C O_{2}$
$3 H_{2} + \frac{3}{2} O_{2} \to 3 H_{2} O$
$2 C O_{2} + 3 H_{2} O \to C_{2} H_{5} O H + 3 O_{2}$
So the Formation enthalpy of ethanol=
$2 \times 394 - \frac{3}{2} \times 568 + 1058 = - 582 K J$

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Similar questions

Q. Given that:

C (s) + O_{2} (g) \to C O_{2} (g)

;

Δ H = - 394

kJ and

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l), Δ H = - 568

kJ and

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

;

Δ H = - 1058

kJ/mole. Using the data, the heat of formation of ethanol is:

Q. Given that,

C_{(s)} + O_{2 (g)} \to C O_{2 (g)};

Δ H = - 394 k J

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)};

Δ H = - 568 k J

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)};

Δ H = - 394 k J

Heat of formation of

C H_{4}

is :

C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 94.3

kcal/mol

C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g); Δ H = - 67.4

kcal/mol

O_{2} (g) \to 2 O (g); Δ H = 117.4

kcal/mol

C O (g) \to C (g) + O (g); Δ H = 230.6

kcal/mol

Calculate

Δ H

for

C (s)

\to

C (g)

in Kcal/mol.

C (s) + O_{2} (g) \to C O_{2}, (g);

Δ H = - 94.3

kcal/mol

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g);

Δ H = - 67.4

kcal/mol

O_{2} (g) \to 2 O (g);

Δ H = 117.4

kcal/mol

C O (g) \to C (g) + O (g);

Δ H = 230.6

kcal/mol.
Calculate

Δ H

for

C (s) \to C (g)

in kcal/mol.

Q. Find the heat of formation of ethyl alcohol from following data

C (s) + O_{2} (g) ⟶ C O_{2} (g)

......

Δ H = - 94 K c a l

H_{2} (g) + I / 2 O_{2} (g) ⟶ H_{2} O (l)

;....

Δ H = - 68 K c a l

C_{2} H_{5} O H (l) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 3 H_{2} O (l)

.....

Δ H = - 327 K c a l

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Given the following:C(s)+O2(g)→CO2(g); ΔH=−394 kJ/mol2H2(g)+O2(g)→2H2O(l); ΔH=−568 kJ/mol C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l); ΔH=−1058 kJ/molUsing the given data, the heat of formation of ethanol is: