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Calculate the Equilibrium Constant for the Following Reaction at 25 \circ

question 39

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Calculate the equilibrium constant for the following reaction at 25 \circ C,
2 IO3-(aq) + 5 Hg(  Calculate the equilibrium constant for the following reaction at 25 <sup> \circ </sup>C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq)  \to  I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  Given the following thermodynamic information. IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup>  \to  I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  ~~~~~~~~ E<sup> \circ </sup> = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  Hg(   )  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ E<sup> \circ </sup>= +0.86 V A)  3  \times  10<sup>-58</sup> B)  6  \times  10<sup>5</sup> C)  3  \times  10<sup>11</sup> D)  6  \times  10<sup>28</sup> E)  3  \times  10<sup>57</sup> ) + 12 H+(aq) \to I2(s) + 5 Hg2+(aq) + 6 H2O(  Calculate the equilibrium constant for the following reaction at 25 <sup> \circ </sup>C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq)  \to  I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  Given the following thermodynamic information. IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup>  \to  I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  ~~~~~~~~ E<sup> \circ </sup> = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  Hg(   )  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ E<sup> \circ </sup>= +0.86 V A)  3  \times  10<sup>-58</sup> B)  6  \times  10<sup>5</sup> C)  3  \times  10<sup>11</sup> D)  6  \times  10<sup>28</sup> E)  3  \times  10<sup>57</sup> )
Given the following thermodynamic information.
IO3-(aq) + 6 H+(aq) + 5 e- \to I2(s) + 3 H2O(  Calculate the equilibrium constant for the following reaction at 25 <sup> \circ </sup>C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq)  \to  I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  Given the following thermodynamic information. IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup>  \to  I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  ~~~~~~~~ E<sup> \circ </sup> = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  Hg(   )  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ E<sup> \circ </sup>= +0.86 V A)  3  \times  10<sup>-58</sup> B)  6  \times  10<sup>5</sup> C)  3  \times  10<sup>11</sup> D)  6  \times  10<sup>28</sup> E)  3  \times  10<sup>57</sup> )         ~~~~~~~~ E \circ = +1.20 V
Hg2+(aq) + 2 e- \to Hg(  Calculate the equilibrium constant for the following reaction at 25 <sup> \circ </sup>C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq)  \to  I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  Given the following thermodynamic information. IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup>  \to  I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  ~~~~~~~~ E<sup> \circ </sup> = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  Hg(   )  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ E<sup> \circ </sup>= +0.86 V A)  3  \times  10<sup>-58</sup> B)  6  \times  10<sup>5</sup> C)  3  \times  10<sup>11</sup> D)  6  \times  10<sup>28</sup> E)  3  \times  10<sup>57</sup> )         ~~~~~~~~        ~~~~~~~~        ~~~~~~~~        ~~~~~~~~ E \circ = +0.86 V


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