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Byju's Answer

Standard XII

Physics

1st Law of Thermodynamics

If the heats ...

Question

If the heats of formation of $C_{2} H_{2}$ and $C_{6} H_{6}$ are 230 KJ mol $^{- 1}$ and 85 KJ mol $^{- 1}$ respectively, the
$Δ$ H value for the trimerisation of $C_{2} H_{2}$ is:

-605 KJ

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-205 KJ

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205 KJ

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605 KJ

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Solution

The correct option is A -605 KJ
(i) $2 C (s) + H_{2} (g) \to C_{2} H_{2}$
(ii) $6 C (s) + 3 H_{2} (g) \to C_{6} H_{6} Δ H = 85 K J$
(iii) $C_{2} H_{2} \to 2 C (s) + H_{2} (g) Δ H = - 230 K J$
(ii) + 3(iii) gives
$3 C_{2} H_{2} \to C_{6} H_{6}$
$Δ H = - 230 \times 3 + 85$
$= - 690 + 85$
$= - 605 K J$
Option A is correct.

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Similar questions

Q. When ethyne is passed through a red hot tube, then formation of benzene takes place:

Δ H_{f (C_{2} H_{2}) (g)}^{\circ} = 230 k J m o l^{- 1}

H_{f (C_{6} H_{6}) (g)}^{\circ} = 85 k J m o l^{- 1}

Calculate the standard heat of trimerisation of ethyne to benzene.

3 C_{2} H_{2} (g) \to C_{6} H_{6} (g)

The molar heat of combustion of $C_{2} H_{2}$ , carbon and hydrogen are $- 1300$ kJ, $- 393.5$ kJ and $- 282.6$ kJ respectively. The enthalpy of $C_{2} H_{2}$ in kJ mol $^{- 1}$ is:

Q. Using the data provided, calculate the multiple bond energy

(k J m o l^{- 1})

of a

C \equiv C

bond in

C_{2} H_{2} .

Given that the heat of formation of

C_{2} H_{2} = 225 k J m o l^{- 1} .

(take the bond energy of

C - H

bond as

350 k J m o l^{- 1}

2 C (s) \to 2 C (g) △ H = 1410 k J m o l^{- 1}

H_{2} (g) \to 2 H (g) △ H = 330 k J m o l^{- 1}

Q. When ethyne is passed through a red hot tube, then formation of benzene takes place:

Δ H_{f (C_{2} H_{2}) (g)}^{\circ} = 230 k J m o l^{- 1}

H_{f (C_{6} H_{6}) (g)}^{\circ} = 85 k J m o l^{- 1}

Calculate the standard heat of trimerisation of ethyne to benzene.

3 C_{2} H_{2} (g) \to C_{6} H_{6} (g)

Q. The enthalpies of formation of

O H (g)

H (g)

and

O (g)

are 21

k J m o l^{- 1}

, 109

k J m o l^{- 1}

and 124

k J m o l^{- 1} .

The value of bond enthalpy

O - H

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If the heats of formation of C2H2 and C6H6 are 230 KJ mol−1 and 85 KJ mol−1 respectively, theΔH value for the trimerisation of C2H2 is:

The correct option is A -605 KJ(i) 2C(s)+H2(g)→C2H2(ii) 6C(s)+3H2(g)→C6H6ΔH=85KJ(iii) C2H2→2C(s)+H2(g)ΔH=−230KJ(ii) + 3(iii) gives3C2H2→C6H6ΔH=−230×3+85=−690+85=−605KJOption A is correct.

If the heats of formation of $C_{2} H_{2}$ and $C_{6} H_{6}$ are 230 KJ mol $^{- 1}$ and 85 KJ mol $^{- 1}$ respectively, the
$Δ$ H value for the trimerisation of $C_{2} H_{2}$ is: