1) The value of K_(eq) for the equilibrium H_(2)(g)+I_(2)(g)leftharpoons 2HI(g) is 794 at 25^circ C What is the value of K_(eq) for the equilibrium below? 1/2H_(2)(g)+1/2I_(2)(g)leftharpoons HI(g) A) 397 B) 0.035 C) 28 D) 1588 E) 0.0013 Answer: C

To determine the value of \( K_{eq} \) for the equilibrium \( \frac{1}{2}H_{2}(g) + \frac{1}{2}I_{2}(g) \rightleftharpoons HI(g) \), we need to understand the relationship between this equilibrium and the given equilibrium \( H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g) \).<br /><br />The given equilibrium is:<br />\[ H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g) \]<br />with \( K_{eq} = 794 \).<br /><br />The equilibrium we need to find the \( K_{eq} \) for is:<br />\[ \frac{1}{2}H_{2}(g) + \frac{1}{2}I_{2}(g) \rightleftharpoons HI(g) \]<br /><br />This equilibrium is essentially half of the original equilibrium. When we halve the coefficients of a reaction, the equilibrium constant \( K_{eq} \) is raised to the power of the reciprocal of the number of times the reaction is halved. In this case, the reaction is halved once (since we are considering half of the original reaction).<br /><br />Therefore, the new equilibrium constant \( K_{eq} \) is:<br />\[ K_{eq, \text{new}} = K_{eq, \text{original}}^{\frac{1}{2}} \]<br /><br />Substituting the given value:<br />\[ K_{eq, \text{new}} = 794^{\frac{1}{2}} \]<br /><br />Calculating the square root of 794:<br />\[text}} = \sqrt{794} \approx 28.]<br /><br />Thus, the value of \( K_{eq} \) for the equilibrium \( \frac{1}{2}H_{2}(g) + \frac{1}{2}I_{2}(g) \rightleftharpoons HI(g) \) is approximately 28.2.<br /><br />Therefore, the correct answerC) 28