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Standard XII

Chemistry

Enthalpy of Combustion

What will be ...

Question

What will be the enthalpy change of conversion of graphite into diamond?
Given $C_{g r a p h i t e}, △_{c o m b} H$ = -391.25 kJ;
$C_{d i a m o n d}, △_{c o m b} H$ = -393.12 kJ

zero

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-391.25 kJ

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-393.12 kJ

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-1.87 kJ

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Solution

The correct option is C -1.87 kJ
$G r a p h i t e ⟶ D i a m o n d Δ H = - [\sum Δ_{C} H (r e a c t a n t s) - \sum Δ_{C} H (p r o d u c t s)]$ $= - [\sum Δ_{C} H (G r a p h i t e) - \sum Δ_{C} H (D i a m o n d)] = - (- 391.25 + 393.12) k J = - 1.87 k J$

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Similar questions

Q. The enthalpies of combustion of

C_{(g r a p h i t e)}

and

C_{(d i a m o n d)}

are

- 393.5

and

- 395.4 k J / m o l

respectively. The enthalpy of conversion of

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

k J / m o l

is:

Q. The enthalpy of transition of C(diamond) to C(graphite) is?
Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

m o l^{- 1}

respectively)

Q. Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of

C_{g r a p h i t e}

and

C_{d i a m o n d}

are -94.05 and -94.50 kcal respectively.

Q. Calculate the enthalpy change accompaning the conversion of

10 g

of graphite into diamond if the heats of combustion of

C (g r a p h i t e)

and

C (d i a m o n d)

are

- 94.05 k c a l

and

- 94.50 k c a l

respectively.

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

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What will be the enthalpy change of conversion of graphite into diamond?Given Cgraphite,△combH = -391.25 kJ; Cdiamond,△combH = -393.12 kJ

The correct option is C -1.87 kJGraphite⟶DiamondΔH=−[∑ΔCH(reactants)−∑ΔCH(products)] =−[∑ΔCH(Graphite)−∑ΔCH(Diamond)]=−(−391.25+393.12)kJ=−1.87kJ

What will be the enthalpy change of conversion of graphite into diamond?
Given $C_{g r a p h i t e}, △_{c o m b} H$ = -391.25 kJ;
$C_{d i a m o n d}, △_{c o m b} H$ = -393.12 kJ

The correct option is C -1.87 kJ
$G r a p h i t e ⟶ D i a m o n d Δ H = - [\sum Δ_{C} H (r e a c t a n t s) - \sum Δ_{C} H (p r o d u c t s)]$ $= - [\sum Δ_{C} H (G r a p h i t e) - \sum Δ_{C} H (D i a m o n d)] = - (- 391.25 + 393.12) k J = - 1.87 k J$