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NISA [10]
3 years ago
12

A reaction is in equilibrium as shown: A + B C + D. Calculate the equilibrium constant in the final concentrations stabilized at

: A= 9.6 M B= 10.0 M C= 4.0 M D= 4.0 M K =
Chemistry
1 answer:
Butoxors [25]3 years ago
8 0

Answer:

K = 0.167

Explanation:

The equilibrium constant, K of a reaction, is defined as the ratio of the concentrations of products and concentrations of reactants.

For the reactions:

A + B ⇄ C + D

For the definition, K is:

K = [C] [D] / [A] [B]

K = [4.0M] [4.0M] / [9.6M] [10.0M]

<h3>K = 0.167</h3>

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
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Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
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Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

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