Assertion- Cgraphite-O2g- CO2g- - H-393-5k-J Cdiamond-O2g- CO2g- - H-395-4k-JReasonR- graphite is more stable than diamond- so heat of combustion is less-

The correct option is A A and R are true and R is correct explanation of A.C (diamond) → C (graphite) nbsp; Δ H= ve = 2 K.J.Thus, graphit ...

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Standard XII

Chemistry

Entropy

Assertion- ...

Question

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,
$Δ H = - 393.5 k . J$
$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,
$Δ H = - 395.4 k . J$
Reason(R): graphite is more stable than diamond, so heat of combustion is less.

A and R are true and R is correct explanation of A.

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A and R are true and R is not correct explanation of A

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A is true, R is false

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A is false, R is true

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Solution

The correct option is A A and R are true and R is correct explanation of A.
C (diamond) $\to$ C (graphite) $Δ H =$ - ve $=$ - 2 K.J.
Thus, graphite is more stable.Heat of combustion is less for stable substance

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Assertion- C(graphite)+O2(g)→CO2(g),ΔH=−393.5k.JC(diamond)+O2(g)→CO2(g),ΔH=−395.4k.JReason(R): graphite is more stable than diamond, so heat of combustion is less.

The correct option is A A and R are true and R is correct explanation of A.C (diamond) → C (graphite) ΔH= - ve = - 2 K.J.Thus, graphite is more stable.Heat of combustion is less for stable substance

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,
$Δ H = - 393.5 k . J$
$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,
$Δ H = - 395.4 k . J$
Reason(R): graphite is more stable than diamond, so heat of combustion is less.

The correct option is A A and R are true and R is correct explanation of A.
C (diamond) $\to$ C (graphite) $Δ H =$ - ve $=$ - 2 K.J.
Thus, graphite is more stable.Heat of combustion is less for stable substance