Given that- Cgraphite- Cg- H-716-7KJ- Cdiamond- Cg- H-714-8KJ- The -H for the following reaction is Cgraphite- Cdiamond

The correct option is A 1.9 KJ(i) C (graphite) nbsp; → C (g) nbsp; Δ H= 716.7 KJC (diamond) nbsp; → C (g) nbsp; Δ H= 714.8 KJ(ii) C ...

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Byju's Answer

Standard XII

Chemistry

Bond Order and Enthalpy

Given that, ...

Question

Given that,
$C_{(g r a p h i t e)} \to C_{(g)}; Δ H = + 716.7 K J .$
$C_{(d i a m o n d)} \to C_{(g)}; Δ H = 714.8 K J .$
The $Δ$ H for the following reaction is
$C_{(g r a p h i t e)} \to C_{(d i a m o n d)}$

1.9 KJ

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-1.9 KJ

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Zero

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714.8 KJ

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Solution

The correct option is A 1.9 KJ
(i) C (graphite) $\to$ C (g) $Δ H =$ 716.7 KJ
C (diamond) $\to$ C (g) $Δ H =$ 714.8 KJ
(ii) C (g) $\to$ C (diamond) $Δ H =$ -714.8 KJ
Adding (i) & (ii) we get
C (graphite) $\to$ C (diamond)
$Δ H =$ 716.7 - 714.8 KJ $=$ - 1.9 KJ

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Given that,C(graphite)→C(g);ΔH=+716.7KJ.C(diamond)→C(g);ΔH=714.8KJ. The ΔH for the following reaction isC(graphite)→C(diamond)

The correct option is A 1.9 KJ(i) C (graphite) → C (g) ΔH= 716.7 KJC (diamond) → C (g) ΔH= 714.8 KJ(ii) C (g) → C (diamond) ΔH= -714.8 KJAdding (i) & (ii) we getC (graphite) → C (diamond)ΔH= 716.7 - 714.8 KJ= - 1.9 KJ

Given that,
$C_{(g r a p h i t e)} \to C_{(g)}; Δ H = + 716.7 K J .$
$C_{(d i a m o n d)} \to C_{(g)}; Δ H = 714.8 K J .$
The $Δ$ H for the following reaction is
$C_{(g r a p h i t e)} \to C_{(d i a m o n d)}$

The correct option is A 1.9 KJ
(i) C (graphite) $\to$ C (g) $Δ H =$ 716.7 KJ
C (diamond) $\to$ C (g) $Δ H =$ 714.8 KJ
(ii) C (g) $\to$ C (diamond) $Δ H =$ -714.8 KJ
Adding (i) & (ii) we get
C (graphite) $\to$ C (diamond)
$Δ H =$ 716.7 - 714.8 KJ $=$ - 1.9 KJ