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Gaseous NOCl decomposes to form the gases NO and Cl2, according to the following equation - VCE - SSCE Chemistry - Question 11 - 2008 - Paper 1

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Gaseous NOCl decomposes to form the gases NO and Cl2, according to the following equation. 2NOCl(g) ⇌ 2NO(g) + Cl2(g) The numerical value of the equilibrium consta... show full transcript

Worked Solution & Example Answer:Gaseous NOCl decomposes to form the gases NO and Cl2, according to the following equation - VCE - SSCE Chemistry - Question 11 - 2008 - Paper 1

Step 1

Determine the relationship between the reactions

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Answer

The provided reaction is:

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

We can write the equilibrium constant expression for this reaction, K1:

K_1 = rac{[NO]^2[Cl_2]}{[NOCl]^2}

The second reaction is:

NO(g) + \frac{1}{2} Cl2(g) ⇌ NOCl(g)

Writing its equilibrium constant expression, K2:

K2=[NOCl][NO][Cl2]1/2K_2 = \frac{[NOCl]}{[NO][Cl_2]^{1/2}}

Step 2

Relate K1 and K2

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To find the relationship between K1 and K2, we note that K2 is derived from K1 by reversing the above reaction and adjusting the stoichiometry:

When we reverse a reaction, the equilibrium constant is the inverse: Kreverse=1KK_{reverse} = \frac{1}{K}

Now, we need to adjust for the coefficients. For K2: K2=1K1K_2 = \frac{1}{\sqrt{K_1}}

Substituting K1: K2=11.6×105K_2 = \frac{1}{\sqrt{1.6 \times 10^{-5}}}

Step 3

Calculate K2

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Calculating K2:

First calculate ( \sqrt{1.6 \times 10^{-5}} ):

\Rightarrow K_2 \approx \frac{1}{0.004} = 250$$ Thus, the numerical value of the equilibrium constant K2 at 35°C is approximately 2.5 × 10².

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