The heat of formation of the compound in the following reaction is H2-Cl2- 2 HCl-44K-cal- Calculate heat of formation of one mole of HCl gas-

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Heat of Formation

The heat of f...

Question

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.

-44 K.cal

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-22 K.cal

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+11 K.cal

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-88 K.cal

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Solution

The correct option is B -22 K.cal
$H_{2} + C l_{2} \to 2 H C l, Δ H = - 44$ K cals

For information of HCl reaction will be

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g), Δ H = - 22$ K cal

Option B is correct.

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Similar questions

The heat of combustion of $C H$ $_{4}$ (g), $C$ (graphite) and $H$ $_{4}$ (g) are respectively $- 20 k . c a l$ , $- 40 k . c a l$ and $- 10 k . c a l$ . The heat of formation of CH $_{4}$ is

H_{2} + C l_{2} \to 2 H C l + 44 K . C a l

. Heat of decomposition of

H C l

is:

The standard heats of formation in K.cal mol $^{- 1}$ of NO $_{2 (g)}$ and N $_{2}$ O $_{4 (g)}$ are 8.0 and 2.0 respectively. The heat of dimerisation of $N O_{2}$ in K.cal is $2 N O_{2 (g)} \to N_{2} O_{4 (g)}$

The heats of neutralisation of acids $A, B, C$ and $D$ with $N a O H$ are -13.5 K.cal, -12.7 K.cal, -11.8 K.cal, -12.4 K.cal respectively. The weakest acid is:

Δ H_{f} =

-98.2 K.Cal/mole

S_{N a} =

36 K.Cal/mole

I_{N a} =

118.5 K.Cal/mole

\frac{1}{2} D_{C l_{2}} =

29 K.Cal/mole

U_{N a C l} =

-184.2 K.Cal/mole
From the data given below for

N a C l

, the electron affinity of chlorine

[- E_{a}]

is:

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The heat of formation of the compound in the following reaction is H2+Cl2→2HCl+44K.cal. Calculate heat of formation of one mole of HCl gas.

The correct option is B -22 K.calH2+Cl2→2HCl,ΔH=−44 K calsFor information of HCl reaction will be12H2(g)+12Cl2(g)→HCl(g),ΔH=−22 K calOption B is correct.

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.

The correct option is B -22 K.cal
$H_{2} + C l_{2} \to 2 H C l, Δ H = - 44$ K cals

For information of HCl reaction will be

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g), Δ H = - 22$ K cal

Option B is correct.