Question

1. the equilibrium constant, $k_c$, is 54.3 at 430 $^circ$c for the reaction below.$ce{h2(g) + i2(g) <=> 2 hi(g)}$if a reaction vessel, initially contained only $ce{h2}$ and $ce{i2}$, each at a concentration of 0.5 m, what is the concentration of $ce{h2}$ at equilibrium?a. 0.107 mb. 0.874 mc. 0.428 md. 0.787 me. 1.57 m24. based on le châteliers principle, in which of the following equilibrium systems will an increase in the pressure (decrease in volume) have no effect on the concentrations of products and reactants?a. $ce{2 h2o(g) + o2(g) <=> 2 h2o2(g)}$b. $ce{n2(g) + 3 h2(g) <=> 2 nh3(g)}$c. $ce{caco3(s) <=> cao(s) + co2(g)}$d. $ce{2 nobr(g) <=> 2 no(g) + br2(g)}$e. $ce{h2(g) + f2(g) <=> 2 hf(g)}$25. for the following exothermic reaction system at equilibrium, which one of the changes below would cause the equilibrium to shift to the right?$ce{br2(g) + 2no(g) <=> 2nobr(g)}$ $delta h^circ$ = -30 kja. increase the volume of the reaction vessel.b. remove some no.c. add some nobr.d. remove some $ce{br2}$.e. decrease the temperature.

Explanation:

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Question 23

Step1: Define change in concentration

Let $x$ = concentration of $\text{H}_2$ that reacts.
Initial concentrations: $[\text{H}_2]_0 = [\text{I}_2]_0 = 0.5\ \text{M}$, $[\text{HI}]_0 = 0\ \text{M}$
Equilibrium concentrations:
$[\text{H}_2] = 0.5 - x$, $[\text{I}_2] = 0.5 - x$, $[\text{HI}] = 2x$

Step2: Write $K_c$ expression

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$
Substitute values:
$$54.3 = \frac{(2x)^2}{(0.5 - x)(0.5 - x)}$$

Step3: Simplify and solve for $x$

Take square root of both sides:
$$\sqrt{54.3} = \frac{2x}{0.5 - x}$$
$$7.369 = \frac{2x}{0.5 - x}$$
$$7.369(0.5 - x) = 2x$$
$$3.6845 - 7.369x = 2x$$
$$3.6845 = 9.369x$$
$$x = \frac{3.6845}{9.369} \approx 0.393\ \text{M}$$

Step4: Calculate $[\text{H}_2]$ at equilibrium

$$[\text{H}_2] = 0.5 - x = 0.5 - 0.393$$

Le Châtelier's principle states that pressure changes affect gas-phase equilibria only if the total moles of gaseous reactants differ from total moles of gaseous products. Reactions with equal moles of gas on both sides are unaffected by pressure changes.

• A: 3 moles gas reactants → 2 moles gas products
• B: 4 moles gas reactants → 2 moles gas products
• C: 0 moles gas reactants → 1 mole gas product
• D: 2 moles gas reactants → 3 moles gas products
• E: 2 moles gas reactants → 2 moles gas products

For the exothermic reaction $\text{Br}_2(g) + 2\text{NO}(g)
ightleftharpoons 2\text{NOBr}(g)\ \ \Delta H^\circ = -30\ \text{kJ}$:

• A: Increasing volume decreases pressure; equilibrium shifts to side with more gas moles (left).
• B: Removing reactant ($\text{NO}$) shifts equilibrium left.
• C: Adding product ($\text{NOBr}$) shifts equilibrium left.
• D: Removing reactant ($\text{Br}_2$) shifts equilibrium left.
• E: For exothermic reactions, decreasing temperature shifts equilibrium to the right (favors heat-releasing direction).

Answer:

A. 0.107 M

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Question 24