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Chemical Reactivity (Group 1)

What is Ht×...

Question

What is $△ H_{t \times n}$ for the decomposition of $1 m o l e$ of $N a C l O_{3}$ ? $△ H_{f} = - 85.7 k c a l / m o l$ for $N a C l O_{3} (s)$ and $△ H_{t} = 98.2 k c a l / m o l$ for $N a C l (s)$

- 183.9 k c a l

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- 91.9 k c a l

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+ 45.3 k c a l

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+ 22.5 k c a l

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- 12.5 k c a l

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Solution

The correct option is E $- 12.5 k c a l$
Given , $N a c l + \frac{3}{2} O_{2} \to N a c l O_{3}$ , $Δ H = - 85.7 k c a l / m o l$ and $Δ H_{f}$ of $N a c l = 98.2 k c a l / m o l$
$Δ H_{r e a c t i o n} = Δ H_{f (p r o d u c t s)} - Δ H_{f (r e a c t a n t s)}$
$- 85.7 = Δ H_{f} (N a c l O_{3}) - Δ H_{f} (N a c l + \frac{3}{2} O_{2})$
$- 85.7 = Δ H_{f} (N a c l O_{3}) - Δ H_{f} (N a c l) - 0$
$Δ H_{f} (N a c l O_{3}) = 98.2 - 85.7 = 12.5 k c a l / m o l$

Suggest Corrections

Similar questions

Δ H^{o}

N a C l O_{3} (s) \to N a C l (s) + \frac{3}{2} O_{2} (g)

(Δ H_{f}^{o}

:

N a C l O_{3} (s) = 358 J / m o l,

N a C l (s) = - 410 j / m o l

O_{2} (g) = 0

k c a l / m o l)

1

Δ H_{r e a c t i o n}

Δ H_{f}^{0}

N a C l O_{3} (s) = - 85.7

/

N a C l (s) = - 98.2

/

O_{2} (g) = 0

/

N a O C l Δ \to N a C l O_{3} + N a C l

C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

3 N a C l O \to 2 N a C l + N a C l O_{3}

4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

106.5

N a C l O_{3}

3 C l_{2} + x N a O H ⟶ N a C l + N a C l O_{3} + H_{2} O

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