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Heat of Formation

Kc values for...

Question

$K_{c}$ values for $\frac{3}{2} H_{2} + \frac{1}{2} N_{2} ⇌ N H_{3}$ are $0.0266$ and $0.0129 a t m^{- 1}$ respectively at $350^{o} C$ and $400^{o} C$ .
Calculate the heat of formation of $N H_{3}$

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Solution

Using van't haff equation $h n (\frac{K_{e}}{K_{1}}) = \frac{Δ^{r} H^{o}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
We have : $T_{1} = 350^{o} C = 623 K, K_{1} = 0.0266 a t m^{- 1}$
$T_{2} = 400^{o} C = 673 K, K_{2} = 0.0129 a t m^{- 1}$
Putting these values in equation, We get:
$ln (\frac{0.0266}{0.0129}) = \frac{Δ^{r} H^{o}}{0.082} [\frac{1}{628} - \frac{1}{678}] \Rightarrow Δ^{r} H^{o} = - 0.497 K J$
Enthalpy of the reaction $= - 0.497 K J$
$∴$ Heat of formation $= 0.497 K J$

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Similar questions

K_{c}

\frac{3}{2} H_{2} + \frac{1}{2} N_{2} ⇌ N H_{3}

are 0.0266 and 0.0129

a t m^{- 1}

respectively at

350^{o} C

400^{o} C

. Calculate heat of formation of

N H_{3}

.

K_{C}

3 / 2 H_{2} + 1 / 2 N_{2} ⇌ {N H}_{3}

0.0129 {a t m}^{- 1}

respectively, at 350

^{o}

^{o}

C. Calculate the heat of formation of

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Q. Heat of formation of

{N H}_{3}

4298 K

and constant pressure is

- 11 k c a l {m o l}^{- 1}

. Calculate heat of formation of

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and constant pressure. What are corresponding values at constant volume?

Δ G^{o} (298 K)

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m o l^{- 1}

. The equilibrium constant

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25^{o} C

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N_{2} + 3 H_{2}_{2} 2 N H_{3}

N H_{3}_{3} \frac{1}{2} N_{2} + \frac{3}{2} H_{2}

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

At equilibrium, the concentration of

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1.2 \times 10^{- 2}, 3.0 \times 10^{- 2}

1.5 \times 10^{- 2} M

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