Equilibrium

Equilibrium

Question

The equilibrium constant for the reaction

SO subscript 3 space left parenthesis straight g space right parenthesis space rightwards harpoon over leftwards harpoon with space space space space space on top space SO subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis
is Kc = 4.9 × 10–2. The value of Kc for the reaction'
2SO2 (g) +O2(g) ⇌ 2SO3 (g) will be

  • 416

  • 2.40 × 10–3

  • 9.8 × 10–2

  • 4.9 × 10–2

Answer

A.

416

straight K apostrophe subscript straight c space equals space open parentheses fraction numerator 1 over denominator 4.9 space straight x space 10 to the power of negative 2 end exponent end fraction close parentheses squared
space equals space fraction numerator 10 to the power of 4 over denominator 4.9 space straight x space 4.9 end fraction space equals space fraction numerator 100 space straight x 100 over denominator 24.01 end fraction
space equals space 4.1649 space space straight x space 100
space equals space 416.49

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