Given- C s - O 2 g - CO 2 g - H -94-2 kcalH 2 g -1-2 O 2 g - H 2 O l - H -68-3 kcalCH 4 g -2 O 2 g - CO 2 g -2 H 2 O l - H -210-8 kcalThe heat of formation of methane in kcal will be -A- -50 kcalB- -20 kcalC- -40 kcalD- -30 kcal

The correct option is B 20 kcalGiven, nbsp;C(s)+O_2(g)→ CO_2(g) Δ H= 94.2 kcal....(1) H_2(g)+1/2O_2(g)→ H_2O(l) Δ H= 68.3 kcal....(2) CH_4(g)+2O_2(g ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy of Combustion

Given, C s + ...

Question

Given,
$C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.2 k c a l$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.3 k c a l$
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.8 k c a l$
The heat of formation of methane in kcal will be :

- 50 k c a l

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- 40 k c a l

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- 30 k c a l

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- 20 k c a l

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Solution

The correct option is D $- 20 k c a l$
Given, $C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.2 k c a l . . . . (1)$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.3 k c a l . . . . (2)$
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.8 k c a l . . . . (3)$
we need $C (s) + 2 H_{2} (g) \to C H_{4} (g)$

multiply equation 2 by 2, we get
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) Δ H = + 2 \times (- 68.3) k c a l . . . . . . . . . . . .4$

reversing the equation 3, we get
$C O_{2} (g) + 2 H_{2} O (l) \to C H_{4} (g) + 2 O_{2} (g) Δ H = (+ 210.8) k c a l . . . . . . . . .5$

adding eqn 1, 4 and 5 to get $C (s) + 2 H_{2} (g) \to C H_{4} (g)$
so, heat of formation of methane $= (- 94.2 + 2 \times (- 68.3) + 210.8) k c a l$
$= - 20 k c a l$

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Similar questions

Q. Given,

C (s) + O_{2} (g) \to C O_{2} (g) Δ H = + 94.2 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 Δ H = k c a l

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = + 210.8 k c a l

The heat of formation of methane in

k c a l

will be:

Q. Given

C (s) + O_{2} (g) \to C O_{2} (g) + 94.2 K c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 K c a l

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) + 210.8 K c a l

The heat of formation of methane in kcal will be:

Q. Using the following thermochemical data,

C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.0 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.0 k c a l

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.0 k c a l

the heat of formation of acetic acaid is :

Heat of formation of $C H_{4}$ is:

If given:

$C (s) + O_{2} (g) \to C O_{2} (g) △ H = - 394 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) △ H = - 284 k J$

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) △ H = - 892 k J$

C (s) + O_{2} (g) \to C O_{2} (g), Δ H = + 94.0 K c a l .

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ H = - 67.7 K c a l .

From the above reactions find how much heat (Kcal mole

^{- 1}

) would be produced in the following reaction:

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

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Given, C(s)+O2(g)→CO2(g) ΔH=−94.2 kcal H2(g)+12O2(g)→H2O(l) ΔH=−68.3 kcal CH4(g)+2O2(g)→CO2(g)+2H2O(l) ΔH=−210.8 kcal The heat of formation of methane in kcal will be :

Given,
$C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.2 k c a l$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.3 k c a l$
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.8 k c a l$
The heat of formation of methane in kcal will be :