H2-Cl2- 2HCl-44 K-Cal- Heat of decomposition of HCl is-

The heat of formation of the compound in the following reaction is $H_{2} + C l_{2} \to 2 H C l + 44 K$ .cal. Calculate heat of formation of one mole of HCl gas.

H_{2} + C l_{2} \to 2 H C l; Δ H = - 44 k c a l

.
Heat of formation of

H C l

is:

At 298 K, the heats of formation of CCl $_{4}$ (g), H $_{2}$ O (g), CO $_{2}$ (g) and HCl (g) are -25, -57, -94 and -22
K.cal mol $^{- 1}$ respectively. The $Δ H$ of the reaction:

$C C l_{4} (g) + 2 H_{2} O \to C O_{2} (g) + 4 H C l (g)$

H_{2} + {C l}_{2} \to 2 H C l

2.02 g

H_{2}

and

35.45 g

{C l}_{2}

are allowed to react, what is the limiting reactant and what mass of

H C l

will be formed?

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H2+Cl2→2HCl+44 K.Cal. Heat of decomposition of HCl is:

The correct option is D +22 K.CalH2+Cl2⟶2HCl+44KcalHeat of decomposition of HCl :-2HCl⟶ΔHdecomp=+44Kcal1HCl⟶ΔHdecomp=+22KcalSo, Heat of decomposition of HCl=+22Kcal

$H_{2} + C l_{2} \to 2 H C l + 44 K . C a l$ . Heat of decomposition of $H C l$ is:

The correct option is D $+ 22 K . C a l$
$H_{2} + C l_{2} ⟶ 2 H C l + 44 K c a l$
Heat of decomposition of $H C l$ :-
$2 H C l ⟶ Δ H_{d e c o m p} = + 44 K c a l$
$1 H C l ⟶ Δ H_{d e c o m p} = + 22 K c a l$
So, Heat of decomposition of $H C l = + 22 K c a l$