In equilibrium: \(\mathrm{SCN}(\mathrm{aq})+\mathrm{Fe}^{+3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right](\mathrm{aq})\) \(\begin{array}{lll}\text { colourless } & \text { yellow } & \text { deep red }\end{array}\) If thiocyanate ions are added in equilibrium mixture (a) The solution becomes colourless. (b) The yellow colour of solution deepens. (c) The red colour of the solution deepens. (d) Concentration of \([\mathrm{Fe}(\mathrm{SCN})]^{2+}\) ion will decrease.

Short Answer

Expert verified
The red color of the solution deepens.

Step by step solution

01

Identify the Direction of Shift

According to Le Chatelier's Principle, when the concentration of one of the reactants is increased, the system will shift to counteract that change, meaning the equilibrium will shift to the right in this case.
02

Predict the Outcome of the Shift

Since the equilibrium shifts to the right, more \[\mathrm{Fe}(\mathrm{SCN})^{2+}\] ions will be formed, hence increasing the concentration of \[\mathrm{Fe}(\mathrm{SCN})^{2+}\] ions.
03

Determine the Color Change

The formation of more \[\mathrm{Fe}(\mathrm{SCN})^{2+}\] ions which are deep red in color will cause the red color of the solution to deepen.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept in chemical equilibrium that describes how a system at equilibrium reacts to disturbances. This principle states that if an external change is applied to a system in equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium. Such changes can include variations in concentration of reactants or products, changes in pressure or volume for gaseous systems, or alterations in temperature.

For example, when the concentration of a reactant is increased, the equilibrium will shift towards the side of the reaction that will use up the added reactant, thereby producing more products. If the concentration of a product is increased, the system will shift the other way. The principle helps us predict the outcomes of changing conditions on the concentrations of reactants and products within a chemical reaction.
Equilibrium Constant
The equilibrium constant, denoted as K, quantifies the balance between the concentrations of reactants and products in a chemical reaction at equilibrium. It is a fixed value at a given temperature and is defined by the concentrations (or pressures for gases) of the reactants and products raised to the power of their coefficients from the balanced equation. For the general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant \(K_{c}\) is given by \(K_{c} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\).

In the context of our exercise, if the concentration of SCN (thiocyanate ions) is increased, according to Le Chatelier's principle, the equilibrium position will shift to the right to form more of the complex iron(III) thiocyanate, \( [Fe(SCN)]^{2+} \), which means that the actual concentration ratio of the products to reactants will change. However, the equilibrium constant itself remains unchanged, as it is only dependent on the temperature.
Complex Ion Formation
Complex ion formation involves the combination of a metal ion with other molecules or ions, often called ligands, resulting in a species known as a complex ion. These complex ions usually have distinct colors based on the metal and ligands involved, offering a visual cue of their presence in a solution. In our exercise, the metal ion \(Fe^{+3}\) reacts with the ligand \(SCN^-\) to form the deep red complex ion \( [Fe(SCN)]^{2+} \).

The formation of a complex ion can significantly alter the equilibrium position of a reaction. Adding more ligands (like SCN-) to the equilibrium mixture leads to an increased formation of the complex ion, which is a vivid example of Le Chatelier's principle at work. The deepening of the red color indicates more complex ion is formed and also showcases the dynamic nature of equilibrium. Complex ion formation is a key concept in coordination chemistry and is exploited in various analytical and industrial processes.

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Most popular questions from this chapter

For the chemical reaction: \(3 \mathrm{X}(\mathrm{g})\) \(+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_{3} \mathrm{Y}(\mathrm{g})\), the amount of \(\mathrm{X}_{3} \mathrm{Y}\) at equilibrium is affected by (a) temperature and pressure (b) temperature only (c) pressure only (d) temperature, pressure and catalyst

A gaseous mixture contains \(0.30\) moles \(\mathrm{CO}, 0.10 \mathrm{moles} \mathrm{H}_{2}\), and \(0.03\) moles \(\mathrm{H}_{2} \mathrm{O}\) vapour and an unknown amount of \(\mathrm{CH}_{4}\) per litre. This mixture is at equilibrium at \(1200 \mathrm{~K}\). \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(K_{\mathrm{C}}=3.9\) What is the concentration of \(\mathrm{CH}_{4}\) in this mixture? The equilibrium constant \(K_{\mathrm{c}}\) equals \(3.92\). (a) \(0.39 \mathrm{M}\) (b) \(0.039 \mathrm{M}\) (c) \(0.78 \mathrm{M}\) (d) \(0.078 \mathrm{M}\)

In a closed tube, \(\mathrm{HI}(\mathrm{g})\) is heated at \(440^{\circ} \mathrm{C}\) up to establishment of equilibrium. If it dissociates into \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{I}_{2}(\mathrm{~g})\) up to \(22 \%\), the dissociation constant is (a) \(0.282\) (b) \(0.0796\) (c) \(0.0199\) (d) \(1.99\)

When \(\alpha\) -D-glucose is dissolved in water, it undergoes a partial conversion to \beta-D-glucose. This conversion, called mutarotation, stops when \(64.0 \%\) of the glucose is in the \(\beta\) -form. Assuming that equilibrium has been attained, what is \(\Delta G^{\circ}\) for the reaction: \(\alpha\) -D-glucose \(\rightleftharpoons \beta-\mathrm{D}\) glucose, at this experimental temperature? (a) \(-R T \log _{10}(1.6)\) (b) \(-R T \log _{10}(1.78)\) (c) \(-R T \log _{e}(1.78)\) (d) \(-R T \log _{e}(1.6)\)

In the equilibrium mixture: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) \(\rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the mole ratio of gases are \(\sqrt{2}: \sqrt{2}: 10\), respectively. What would be the effect on the mole ratio on adding 5 mole of He gas at constant pressure? (a) No change (b) The new molar ratio becomes \(1: 1: 5\) (c) The new molar ratio becomes \(2: 2: 5\) (d) The new molar ratio becomes \(2: 2: 5 \sqrt{2}\)

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