Equilibrium constant $K_{p}$ for $2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ is $0.0118$ atm at $1065^{o} C$ and heat of dissociation is $42.4 k C a l$ . Find equilibrium constant at $1132^{o} C$

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Solution

$\frac{T_{1}}{T_{2}} = L n \frac{K_{2}}{K_{1}}$

$e^{2.58} = \frac{K_{2}}{K_{1}}$
k1 = 0.0118

$K_{2} = 2.58 \times 0.0118$
$= 0.030$

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Equilibrium constant Kp for 2H2S(g)⇌2H2(g)+S2(g) is 0.0118 atm at 1065oC and heat of dissociation is 42.4kCal. Find equilibrium constant at 1132oC

T1T2=LnK2K1e2.58=K2K1k1 = 0.0118K2=2.58×0.0118 =0.030

Equilibrium constant $K_{p}$ for $2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ is $0.0118$ atm at $1065^{o} C$ and heat of dissociation is $42.4 k C a l$ . Find equilibrium constant at $1132^{o} C$

$\frac{T_{1}}{T_{2}} = L n \frac{K_{2}}{K_{1}}$

$e^{2.58} = \frac{K_{2}}{K_{1}}$
k1 = 0.0118

$K_{2} = 2.58 \times 0.0118$
$= 0.030$