For the change, $C_{d i a m o n d} ⟶ + C_{g r a p h i t e}; Δ H = - 1.89 k J$ , if 6 g of diamond and 6 g of graphite are separately burnt to yield $C O_{2}$ the heat liberated in first case is:

less than in the second case by 1.89 kJ

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less than in the second case by 11.34 kJ

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less than in the second case by 14.34 kJ

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more than in the second case by 0.945 kJ

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Solution

The correct option is D more than in the second case by 0.945 kJ
$C_{D} ⟶ C_{G}; Δ H = - 1.89$ ...( i)
$C_{D} + O_{2} ⟶ C O_{2}; Δ H = - q_{1}$ ...( ii)
$C_{G} + O_{2} ⟶ C O_{2}; Δ H = - q_{2}$ ...( iii)
Subtract equation (iii) from equation (ii).
$C_{D} ⟶ C_{G}; Δ H = - q_{1} + q_{2}$ ...(iv)
From equations (i) and (iv)
$q_{1} + q_{2} = - 1.89$
or $q_{2} - q_{1} = - 1.89 f o r 12 g C_{D \to G}$
Thus, $f o r 6 g C_{D \to G}$
$q_{2} - q_{1} = \frac{- 1.89}{2} = - 0 .945 k J$

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Similar questions

C_{d i a m o n d} ⟶ C_{g r a p h i t e}; Δ H = - 1.89 k J

6 g

6 g

C O_{2}

C (g r a p h i t e) \to C (d i a m o n d), δ H = 1.9 k J .

C O_{2},

C (g r a p h i t e) \to C (d i a m o n d), △ H = 1.9 k J .

6

6

C O_{2}

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