Heat of combustion Δ H^ for C(s), H_2(g) and CH_4(g) are -94,-68 and -213 kcal mol^ -1 , then, Δ H^ for C(s)+2 H_2(g) arrow CH_4(g) is:

Correct answer: A. -17 kcal (i) C_ (s) +O_ 2(~g) arrow CO_ 2(g) ; Δ H_i=-94 kcal / mole (ii) 2 H_ 2(g) +O_ 2(g) arrow 2 H_2 O_ (l) ; Δ H_ ii =-68 × 2 kcal / mole (iii) CH_ 4(g) +2 O_ 2(g) arrow CO_ 2(g) + 2 H_2 O_ (l) ; Δ H_ iii =-213 kcal / mol (iv) C_ (s) +2 H_ 2(g) arrow CH_ 4(~g) ; Δ H_ IV = ? By applying Hess's law we can compute Δ H_ iv . aligned & Δ H_ iv =Δ H_ i +Δ H_ ii -Δ H_ iii & =(-94-68 × 2+213) kcal & =-17 kcal aligned

NEET 2002

Exam: NEET
Year: 2002
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2002 · Chemistry · Thermodynamics (C)

Heat of combustion Δ H^ for C(s), H_2(g) and CH_4(g) are -94,-68...

Heat of combustion

\Delta \mathrm{H}^{\circ}

for

\mathrm{C}(s), \mathrm{H}_2(g)

and

\mathrm{CH}_4(g)

are

-94,-68

and

-213 \mathrm{kcal}

\mathrm{mol}^{-1}

, then,

\Delta \mathrm{H}^{\circ}

for

\mathrm{C}(s)+2 \mathrm{H}_2(g) \rightarrow

\mathrm{CH}_4(g)

is:

Options

A $-17 \mathrm{kcal}$
B $-111 \mathrm{kcal}$
C $-170 \mathrm{kcal}$
D $-85 \mathrm{kcal}$

Correct answer

A. $-17 \mathrm{kcal}$

Step-by-step solution

(i)

\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(g)}

;

\Delta \mathrm{H}_i=-94 \mathrm{kcal} / \mathrm{mole}

(ii)

2 \mathrm{H}_{2(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}

;

\Delta \mathrm{H}_{\mathrm{ii}}=-68 \times 2 \mathrm{kcal} / \mathrm{mole}

(iii)

\mathrm{CH}_{4(g)}+2 \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}+ 2 \mathrm{H}_2 \mathrm{O}_{(l)} ; \Delta \mathrm{H}_{\mathrm{iii}}=-213 \mathrm{kcal} / \mathrm{mol}

(iv)

\mathrm{C}_{(s)}+2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(\mathrm{~g})}

;

\Delta \mathrm{H}_{\mathrm{IV}}=

? By applying Hess's law we can compute

\Delta \mathrm{H}_{\mathrm{iv}}

\begin{aligned} & \therefore \Delta \mathrm{H}_{\mathrm{iv}}=\Delta \mathrm{H}_{\mathrm{i}}+\Delta \mathrm{H}_{\mathrm{ii}}-\Delta \mathrm{H}_{\mathrm{iii}} \\ & =(-94-68 \times 2+213) \mathrm{kcal} \\ & =-17 \mathrm{kcal} \end{aligned}

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