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Given the equilibrium, $$A_2(g) + 4C(g) \rightleftharpoons 2AC_2(g),\ K_1 = 4.8$$ It follows that, for the reaction, $$AC_2(g) \rightleftharpoons \frac{1}{2}A_2(g) + 2C(g),\ K_2 = X$$ X would be A - VCE - SSCE Chemistry - Question 8 - 2002 - Paper 1

Question 8

$Given-the-equilibrium,--$$A_2(g)-+-4C(g)-\rightleftharpoons-2AC_2(g),\-K_1-=-4.8$$--It-follows-that,-for-the-reaction,--$$AC_2(g)-\rightleftharpoons-\frac{1}{2}A_2(g)-+-2C(g),\-K_2-=-X$$--X-would-be--A-VCE-SSCE Chemistry-Question 8-2002-Paper 1.png$

Given the equilibrium, $$A_2(g) + 4C(g) \rightleftharpoons 2AC_2(g),\ K_1 = 4.8$$ It follows that, for the reaction, $$AC_2(g) \rightleftharpoons \frac{1}{2}A_2(g... show full transcript

Worked Solution & Example Answer:Given the equilibrium, $$A_2(g) + 4C(g) \rightleftharpoons 2AC_2(g),\ K_1 = 4.8$$ It follows that, for the reaction, $$AC_2(g) \rightleftharpoons \frac{1}{2}A_2(g) + 2C(g),\ K_2 = X$$ X would be A - VCE - SSCE Chemistry - Question 8 - 2002 - Paper 1

Step 1

Finding the Reverse Reaction Equilibrium Constant

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Answer

Given the equilibrium constant for the forward reaction is ( K_1 = 4.8 ), we can determine the equilibrium constant for the reverse reaction. The relationship between the equilibrium constants of forward and reverse reactions is given by: $K_{reverse} = \frac{1}{K_{forward}}$ Thus, $K_2 = \frac{1}{K_1} = \frac{1}{4.8}$

Step 2

Final Answer Extraction

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Answer

Therefore, X = ( \frac{1}{4.8} ), which corresponds to option A.

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