The equilibrium constants for the reaction- N 2 g- O 2 g- 2NOg at 1727 o C and 2227 o C are 4-08- 10 -4 and 3-6-10-3 respectively- Calculate the enthalpy change for the reaction- Given R-1-987 cal mol-1 K-1

We know the relation between enthalpy change and potential #160; Δ S = Δ nEf Δ T = 3.1 × 10^ 3500 6.01 × 10^ 6 this is answer ...

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Byju's Answer

Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

The equilibri...

Question

The equilibrium constants for the reaction, $N_{2} (g) + O_{2} (g) ⇋ 2 N O (g)$ at $1727^{o} C$ and $2227^{o} C$ are $4.08 \times 10^{- 4}$ and $3.6 \times 10^{- 3}$ respectively. Calculate the enthalpy change for the reaction. (Given $R = 1.987 c a l m o l^{- 1} K^{- 1}$ )

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Solution

we know the relation between enthalpy change and potential
$Δ S = \frac{Δ n E f}{Δ T}$

= $\frac{3.1 \times 10^{- 3}}{500}$

$6.01 \times 10^{- 6}$
this is answer

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Similar questions

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The equilibrium constants for the reaction, N2(g)+O2(g)⇋2NO(g) at 1727oC and 2227oC are 4.08×10−4 and 3.6×10−3 respectively. Calculate the enthalpy change for the reaction. (Given R=1.987 cal mol−1K−1)

we know the relation between enthalpy change and potential ΔS=ΔnEfΔT=3.1×10−35006.01×10−6this is answer

The equilibrium constants for the reaction, $N_{2} (g) + O_{2} (g) ⇋ 2 N O (g)$ at $1727^{o} C$ and $2227^{o} C$ are $4.08 \times 10^{- 4}$ and $3.6 \times 10^{- 3}$ respectively. Calculate the enthalpy change for the reaction. (Given $R = 1.987 c a l m o l^{- 1} K^{- 1}$ )

we know the relation between enthalpy change and potential
$Δ S = \frac{Δ n E f}{Δ T}$

= $\frac{3.1 \times 10^{- 3}}{500}$

$6.01 \times 10^{- 6}$
this is answer