The heat change during the reaction $24 g C$ and $128 g S$ following the change $C + S_{2} \to C S_{2}; Δ H = 22 K c a l$

22 K c a l

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11 K c a l

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44 K c a l

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32 K c a l

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Solution

The correct option is C $44 K c a l$
When 1 mole of $C$ reacts with one mole of $S_{2}$ ) (i.e 12 grams of carbon reacted with 64 grams of sulfur),The heat energy change is $22 k . c a l$
If 2 mole of $C$ reacts with 2 mole of $S_{2}$ ) (i.e 24 grams of carbon reacted with 128 grams of sulfur),The heat energy change will also be doubled as the carbon and sulfur.So the heat energy change is $44 k . c a l$ .
Hence option C is correct.

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Similar questions

24

128

C + S_{2} \to C S_{2}

Δ

= 22

C_{D i a m o n d} + O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

C_{G r a p h i t e} + O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

C_{d i a m o n d} \to C_{g r a p h i t e}

C (s) + 2 S (s) \to {C S}_{2}

+ O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

+ O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

\to

For the following reaction

$C_{d i a m o n d} + O_{2} \to C O_{2} (g)$ ; $Δ H = - 94.3 k c a l$

$C_{g r a p h i t e} + O_{2} \to C O_{2} (g)$ ; $Δ H = - 97.6 k c a l$

The heat required to change 1 g of $C_{d i a m o n d} \to C_{g r a p h i t e}$ is

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