For the reaction: $C_{2} H_{6} \to_{}^{} C_{2} H_{4} + H_{2}$ . The reaction enthalpy $∆ rH =$ __________ $kJ {mol}^{- 1}$ . (Round off to the Nearest Integer).
[Given: Bond enthalpies in $kJ {mol}^{- 1}; C - C : 347; C = C : 611; C - H : 414; H - H : 436$ ]

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Solution

The above question can be solved as:
Step 1:
$C_{2} H_{6} \to C_{2} H_{4} + H_{2} ΔH = ? ?$
Step 2:
$ΔH = [(E_{C - C} + 6 E_{C - H}) - (E_{C = C} + 4 E_{C - H} + E_{H - H}) ΔH = [347 + 6 \times 414] - (611 + 4 \times 414 + 433] = 2831 - 2700 = 131 kJ / mol$
Hence, the reaction enthalpy $∆ rH$ is $131 kJ {mol}^{- 1}$

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Similar questions

Q. For the reaction

C_{2} H_{6} \to C_{2} H_{4} + H_{2}

the reaction enthalpy,

Δ_{r} H

_____

k J m o l^{- 1}

.
[Round off to the Nearest Integer]
[Given : Bond enthalpies in

k J m o l^{- 1}

C - C : 347, C = C : 611;
C - H : 414, H - H : 436]

Complete and balance the following equations

(ⅰ ) $N_{2}$ + $H_{2}$ $\to$ $. . . . .$

(ⅱ) ${NH}_{3}$ + ${Cl}_{2} E x c e s s$ $\to$ $. . . . .$

(ⅱⅰ) $PbO$ + ${NH}_{3}$ $\to$ $. . . . . + . . . . + H_{2} O$

(ⅳ) ${NH}_{3}$ + $O_{2}$ $\overset{pt}{\to} 800 ° C$ $. . . . . + . . . . . .$

Consider the below-given reaction. The percentage yield of an amide product is ______. (Round off to the Nearest Integer).

[Given: Atomic mass: $C$ $= 12.0 u$ , $H$ $= 1.0 u$ , $N$ $= 14.0 u$ , $O$ $= 16.0 u$ , $Cl$ $= 35.5 u$ ]

The amount of solutes ‘A’, ‘B’ and ‘C’ in $300$ g of water at $25 ° C, 40 ° C$ and $60 ° C$ in their saturated solutions are given below. Identify the correct statement from the following when hot saturated solutions of A, B and C are cooled slowly from $60 ° C$ to $25 ° C$ .

	$25 ° C$	$40 ° C$	$60 ° C$
A	$110 g$	$90 g$	$105 g$
B	$140 g$	$170 g$	$180 g$
C	$130 g$	$150 g$	$140 g$

The amounts of solutes ‘A’, 'B' and ‘C’ in $300$ g of water at $25 ° C$ , $40 ° C$ and $60 ° C$ in their saturated solutions are given below. Identify the true statement among the following:

	$25 ° C$	$40 ° C$	$60 ° C$
A	$110$ g	$90$ g	$105$ g
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C	$130$ g	$150$ g	$140$ g

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For the reaction: C2H6→C2H4+H2. The reaction enthalpy ∆rH= __________kJmol-1. (Round off to the Nearest Integer).[Given: Bond enthalpies in kJmol-1;C-C:347;C=C:611;C-H:414;H-H:436]

The above question can be solved as:Step 1:C2H6→C2H4+H2ΔH=??Step 2:ΔH=[(EC-C+6EC-H)-(EC=C+4EC-H+EH-H)ΔH=[347+6×414]-(611+4×414+433]=2831-2700=131kJ/molHence, the reaction enthalpy ∆rHis 131kJmol-1

For the reaction: $C_{2} H_{6} \to_{}^{} C_{2} H_{4} + H_{2}$ . The reaction enthalpy $∆ rH =$ __________ $kJ {mol}^{- 1}$ . (Round off to the Nearest Integer).
[Given: Bond enthalpies in $kJ {mol}^{- 1}; C - C : 347; C = C : 611; C - H : 414; H - H : 436$ ]