SATHEE: Chemical and Ionic Equilibrium 1 Question 20

Chemical and Ionic Equilibrium 1 Question 20

21. For the reversible reaction,

$$ \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$

at $500^{\circ} C$ , the value of $K_{p}$ is $1.44 \times 10^{- 5}$ when partial pressure is measured in atmosphere. The corresponding value of $K_{c}$ with concentration in $mol / L$ is

$(2000, 5, 1 M)$

(a) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 500)^{- 2}}$

(b) $\frac{1.44 \times 10^{- 5}}{(8.314 \times 773)^{- 2}}$

(d) $\frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}}$

Show Answer

Solution:

$\mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$

$Δ n = - 2$

$\begin{aligned} K_{p} & = K_{c} (R T)^{Δ n} K_{c} & = \frac{K_{p}}{(R T)^{Δ n}} = \frac{1.44 \times 10^{- 5}}{(0.082 \times 773)^{- 2}} \end{aligned}$

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Chemical and Ionic Equilibrium 1 Question 20

21. For the reversible reaction,

Solution:

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