The equilibrium constant (K.c) for the reaction 2HCl(g)⇌H2(g)+Cl2(g) is 4×10-34 at 25°C. What is the equilibrium constant for the reaction? 12H2(g)+12Cl2(g)⇌HCl(g)

2HCl(g)⇌H2(g)+Cl2(g) Kc = 4×10-3412H2 + 12Cl2→HCl Kc1=1Kc =14×10-34Kc1=12×10-17=0.5×1017 =5×1016

The equilibrium constant (K.c) for the reaction 2HCl(g)⇌H2(g)+Cl2(g) is 4×10-34 at 25°C. What is the equilibrium constant for the reaction? 12H2(g)+12Cl2(g)⇌HCl(g)

2HCl(g)⇌H2(g)+Cl2(g) Kc = 4×10-3412H2 + 12Cl2→HCl Kc1=1Kc =14×10-34Kc1=12×10-17=0.5×1017 =5×1016

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The equilibrium constant (K._c) for the reaction $2 HCl (g) ⇌ H_{2} (g) + {Cl}_{2} (g)$ is $4 \times 10^{- 34} at 25 ° C .$ What is the equilibrium constant for the reaction? $\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) ⇌ HCl (g)$

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$2.5 \times 10^{33}$

None of these

$5 \times 10^{16}$

$2 \times 10^{- 17}$

answer is C.

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Detailed Solution

2HCl(g)⇌H2(g)+Cl2(g) K_c= 4 $\times$ 10^-34

$\frac{1}{2}$ H₂ + $\frac{1}{2}$ Cl₂ $\to$ HCl $K_{c}^{1}$ = $\frac{1}{\sqrt{K_{c}}}$ = $\frac{1}{\sqrt{4 \times 10^{- 34}}}$

$K_{c}^{1}$ = $\frac{1}{2 \times 10^{- 17}}$ = $0.5 \times 10^{17} = 5 \times 10^{16}$

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The equilibrium constant (K.c) for the reaction 2HCl(g)⇌H2(g)+Cl2(g) is 4×10-34 at 25°C. What is the equilibrium constant for the reaction? 12H2(g)+12Cl2(g)⇌HCl(g)

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The equilibrium constant (K._c) for the reaction $2 HCl (g) ⇌ H_{2} (g) + {Cl}_{2} (g)$ is $4 \times 10^{- 34} at 25 ° C .$ What is the equilibrium constant for the reaction? $\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) ⇌ HCl (g)$