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The enthalpy ...

Question

The enthalpy changes of the following reactions at $27^{\circ} C$ are
$N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)$ $Δ_{r} H = - 411 k J / m o l$
$H_{2} (g) + S (s) + 2 O_{2} (g) \to H_{2} S O_{4} (l)$ $Δ_{r} H = - 811 k J / m o l$
$2 N a (s) + S (s) + 2 O_{2} (g) \to N a_{2} S O_{4} (s)$ $Δ_{r} H = - 1382 k J / m o l$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$ $Δ_{r} H = - 92 k J / m o l;$
from these data, the heat change of reaction at constant volume (in kJ/mol) at $27^{\circ} C$ for the process $2 N a C l (s) + H_{2} S O_{4} (l) \to N a_{2} S O_{4} (s) + 2 H C l (g)$

A

67

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B

62.02

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C

71.98

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D

None

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Solution

The correct option is A 67
$Δ H_{r e a c t i o n} =$ $2 Δ H_{f} H C l$ + $Δ H_{f} N a_{2} S O_{4}$ $- Δ H_{f} N a_{2} S O_{4}$ $- 2 Δ H_{f} N a C l$
$Δ H_{r e a c t i o n} =$ **2(-92)+(-1382)-(--811)-(-2411)
$Δ H_{r e a c t i o n} =$ = 67kJ/mol**

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Similar questions

Q. From the following data at

25^{\circ} C

.

Reaction	$Δ_{r} H^{\circ}$ kJ/mol
$\frac{1}{2} H_{2} (g) + \to \frac{1}{2} O_{2} (g)$	$+ 42$
$H_{2} (g) + \frac{1}{2} O_{2} (g) H_{2} O (g)$	$- 242$
$H_{2} (g) \to 2 H (g)$ $O_{2} (g) \to 2 O (g)$	$+ 436$ $+ 495$

$C (s) + O_{2} (g) \to C O_{2} (g), Δ_{r} H = - 94.3 k c a l / m o l$

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ_{r} H = - 67.4 k c a l / m o l$

$O_{2} (g) \to 2 O (g), Δ_{r} H = 117.4 K c a l / m o l$

$C O (g) \to C (g) + O (g), Δ_{r} H = 230.6 K c a l / m o l$

Calculate: $Δ_{r} H f o r C (s) \to C (g) i n K c a l / m o l$

Q. Calculate the enthalpy of formation of ethyl alcohol from the following data:

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

Δ_{r} H^{\circ} = - 1368.0 k J

C (s) + O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{\circ} = - 393.5 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

Δ_{r} H^{\circ} = - 286.0 k J

Δ_{r} H

Δ_{r} S

for the reaction

M g O (s) + C (s) \to M g (s) + C O (g)

491.18

m o l^{- 1}

197.67

m o l^{- 1}

respectively. Calculate the temperature at which Gibb's energy change will be zero.

Q. Calculate the enthalpy of formation of methane from the following data:

C (s) + O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{\circ} = - 393.5 k J

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)

Δ_{r} H^{\circ} = - 571.8 k J

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

Δ_{r} H^{\circ} = - 890.3 k J

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