Fake Blood and Equilibrium
- Page ID
- 418922
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This page will cover the following concept(s) from the ACS Examinations Institute General Chemistry ACCM:
- VI.H.2. Gibb's free energy is a state function that simultaneously calculates entropy for the system and surroundings, and is useful for determining whether or not a process occurs spontaneously
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VIII. A. Both physical and chemical changes may occur in either direction (e.g., from reactants to products, or products to reactants).
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VIII. B. When opposing processes both occur at the same rate, the net change is zero.
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VIII. C. For chemical and physical processes, the equilibrium state can be characterized via the equilibrium constant.
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VIII. F. F. Thermodynamics provides mathematical tools to understand equilibrium systems quantitatively.
Introduction
In movies of a variety of genres, there is often a necessity for fake blood—something that not only mimics the color of the blood but the viscosity as well. To achieve this using practical effects, there is also another stipulation: the fake blood must immediately appear out of the blue after the stimulus, similar to how we immediately start bleeding after a traumatic event. The equilibrium reaction between potassium thiocyanate and iron nitrate accomplishes all of these things and has been used for decades in movies as fake blood.1
\[Fe^{3+}+SCN^- ⇋ Fe(SCN)^{2+}\]
Figure \(\PageIndex{1}\). Bloody Handed Woman in "The Texas Chainsaw Massacre" movie (Christopher Ross/A24)
Equilibrium
To first understand the fundamentals behind the fake blood reaction, an understanding of equilibrium reactions is necessary.2 What does it even mean for a reaction to be at equilibrium? If a reaction is at equilibrium, the rate of the formation of products is equal to the rate of depletion of products to form reactants. In other words, the rate of the forward reaction is equal to the rate of the reverse reaction. As a result, there is no further net change in the concentrations of the reactants and the products.
A key concept of equilibrium that must be explored is the value K.3 K is known as the equilibrium constant and represents the ratio of products versus reactants when the reaction is at equilibrium.
\[K=\dfrac{[Products]}{[Reactants]}\]
Analyzing the magnitude of K reveals whether a reaction will produce more reactants or products once the reaction reaches equilibrium.
The common guideline is if K is much greater than 1, the products of the reaction are favored at equilibrium and ΔG° (Do not worry about ΔG° right now, you will learn about it later down the page.) is less than 0. This indicates at equilibrium more products will be formed than reactants. Looking back at the equation, the concentration of products will greatly outweigh the concentration of reactants. If K is much smaller than 1, then the reactants of the reaction are favored at equilibrium and ΔG° is greater than 0. This indicates at equilibrium more reactants will be formed than products. Looking back at the equation, the concentration of reactants will greatly outweigh the concentration of products. If K equals 1, then neither the products nor reactants are favored at equilibrium. In this sense, all reactions are ‘equilibrium reactions’. No matter the K value, there will be some reaction happening — it’s just a question of the relative concentration of reactants versus products.
For clarification, the equilibrium constant, K, has no relationship with the speed of the reaction. A chemical reaction may take ten seconds or ten years to reach equilibrium. K is purely driven by thermodynamics and can be used to predict the ratio of products to reactants after the reaction has reached equilibrium.
In the case of the reaction between Potassium Thiocyanate and Iron Nitrate, K is much greater than 1. As a result, after the reactants react with the products, the concentration of Fe(SCN)2+ will be much greater than Fe3+ and SCN-, creating the appearance of fake blood.
Free Energy
Now, we will introduce another concept: ∆G. ∆G is defined as the ‘change in free energy’ of a reaction.4 To put this term into context, think about a chemical reaction as a system. G is the energy available in the system to do work. Every chemical reaction has reactants and products, and these reactants and products have a section of that available energy that they can use. ∆G specifically deals with how this free energy changes in the system as the reaction happens. A reaction can move in such a way that the free energy of the reactants is lower than that of the products, or vice-versa. The free energy between the reactants and products can even be exactly equal! When this occurs in an equation, we would say the equation is at equilibrium. Intuitively, this makes sense—when there is no change in free energy, there is no net impact on the energy of the overall system.
The main application of ΔG is to predict whether or not a reaction will happen spontaneously. If ΔG < 0, then the reaction will proceed spontaneously in the forward direction (reactants -> products). As a result, reactants will be depleted and products will be formed. If ΔG = 0, then the reaction has reached equilibrium, and no net change in reactants or products. If ΔG > 0, then the reaction will proceed spontaneously in the reverse direction (products -> reactants). As a result, products will be depleted and reactants will be formed. When the reaction is said to be spontaneous, this does not mean that the reaction will happen in an instant. Rather, a spontaneous reaction will occur without any input of free energy. This may be in ten seconds or ten years.
In order to produce fake blood, the reaction must be spontaneous. We do not want to add any free energy (like a battery) in order for this reaction to take place. Therefore, the ΔG of the reaction between potassium thiocyanate and iron nitrate will have a ΔG < 0.
How exactly can we calculate the free energy of a specific reaction? With the help of one particular equation, we can use the information that we are given in order to find the free energy available to a certain reaction.
\[ΔG = ΔG° + RTlnQ\]
Do not be intimidated by this equation! Let’s break down each part. We already know that ΔG is the free energy of the reaction, ΔG° looks similar to ΔG but do not be fooled! The little circle on ΔG° represents standard conditions. As a result, this value will be different compared to ΔG. R is a gas constant and will always be 8.314J/mol·K. T represents the temperature measured in the unit kelvin. Last but most definitely not least, we have Q which is the reaction quotient. Q is very similar to the value K referenced above in the previous section. Like K, Q represents the ratio of products to reactants. However, unlike K, Q is used to represent the ratio of products to reactants at any other point in the reaction. In comparison, K can only be used when the reaction is at a state of equilibrium.
Comparing Q and K is very useful to determine the direction the reaction will proceed in. If Q < K, the reaction will proceed in the forward direction and reactants will be depleted while products will be produced. If Q > K, the reaction will proceed in the reverse direction and products will be depleted and reactants will be produced. If the value of Q is analyzed when the reaction is at equilibrium, Q is equal to K and there is no net change in products and reactants! Therefore, when Q = K, the reaction is at equilibrium. Thus, under standard conditions, we can use another equation to help us find ΔG° at equilibrium.
\[ΔG°=-RTlnK\]
Standard Conditions: 273 K or 0° C, 1 atm
Let's apply this concept of ΔG, Q, and K to the reaction between Potassium Thiocyanate and Iron Nitrate! Let's say an actor coats his or her hand with iron nitrate and a knife with ammonium thiocyanate. As the two objects come into contact with each other, no product (fake blood) has been created yet. Therefore, there are more reactants than products and Q is less than 1. We know K is going to be greater than 1 due to the nature of the reaction. Since Q is less than K, the reaction will move spontaneously in the forward direction and ΔG will be negative. Reactants will be depleted and products (fake blood) will be produced.
Fake Blood Reaction
So now that you have a good overview of reaction equilibria and free energy, let’s look at the fake blood reaction!
In movies, when actors want to emulate the appearance of fake blood, they coat their bodies with colorless iron nitrate. Then, they would coat a different material — say a sword, for example — with ammonium thiocyanate. Upon contact between the two, an equilibrium reaction immediately occurs, and fake blood is created!
Let’s take a look at this more deeply with a sample problem based on the concepts explained above.
An actor's hand is coated with Iron(III) Nitrate and a knife is coated with Ammonium Thiocyanate. The blunt knife slices the actor's hand and fake blood is created once the reaction has reached equilibrium.
Find ΔG° given the equation and final concentrations below.
\[Fe^{3+}+SCN^- ⇋ Fe(SCN)^{2+}\]
\[[SCN^-] = 0.0064M\]
\[[Fe^{3+}] = 0.0078 M\]
\[[Fe(SCN)^{2+}] = 0.0065 M\]
- Answer
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This problem gives you the concentrations of SCN, Fe, and Fe(SCN) and asks you to find the value of ΔG°. In order to find ΔG°, you must use the equation given below. Given the concentrations of the products and reactants, you can calculate the value of K by taking the concentrations of the products and dividing them by the concentrations of the reactants.
\[K_c=\dfrac{[Fe(SCN)^{2+}]}{[Fe^{3+}][SCN^-]}\]
Therefore, the value of K would be:
\[K_c=\dfrac{[0.0065]}{[0.0078][0.0064]}\]
\[K_c=130\]
After finding the equilibrium constant, we can now use the equation
\[ΔG°=-RTlnK\]
R, the gas constant, is known as 8.314 J/mol·K while the temperature can be assumed to be at room temperature at 298K. Because the reaction is at equilibrium, ΔG will be 0. Therefore, we can plug these values into the equation.
\[ΔG°=-(\dfrac{8.314 J}{mol·K})(298K)ln(130)\]
Now that we know all the values needed to solve for ΔG°, we can simply calculate our value which will be −12060 J/mol or -12.06 kJ/mol.
Since the ΔG° of the reaction is negative, we can interpret that the reaction will occur spontaneously. This calculation supports our observations and conclusions we came to throughout the discussion about equilibrium and free energy!
Let's do a follow up problem that incorporates the concept of Q! Use the values from the previous question to answer this one, including the value for ΔG°!
Find the value of ΔG with the initial concentrations of reactants and products given below. After calculating the value of ΔG, predict whether the reaction will proceed in the forward or reverse direction by analyzing the value of ΔG and comparing the values of K and Q.
\[Fe^{3+}+SCN^- ⇋ Fe(SCN)^{2+}\]
\[[SCN^-] = 0.0071M\]
\[[Fe^{3+}] = 0.0092 M\]
\[[Fe(SCN)^{2+}] = 0.0045 M\]
- Answer
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This problem gives you the concentrations of SCN, Fe, and Fe(SCN) and asks you to find the value of ΔG. In order to find ΔG, you must use the equation given below. Given the concentrations of the products and reactants, you can calculate the value of Q by taking the concentrations of the products and dividing them by the concentrations of the reactants.
\[Q=\dfrac{[Fe(SCN)^{2+}]}{[Fe^{3+}][SCN^-]}\]
Therefore, the value of Q would be:
\[Q=\dfrac{[0.0045]}{[0.0092][0.0071]}\]
\[Q=68.89\]
After finding the equilibrium constant, we can now use the equation
\[ΔG = ΔG° + RTlnQ\]
R, the gas constant, is known as 8.314 J/mol·K while the temperature can be assumed to be at room temperature at 298K. Using the values from the last problem, ΔG° = -12.06. Therefore, we can plug these values into the equation.
\[ΔG=−12060 J/mol+(\dfrac{8.314 J}{mol·K})(298K)ln(68.89)\]
Now that we know all the values needed to solve for ΔG, we can simply calculate our value which will be −1570 J/mol or -1.57 kJ/mol.
ΔG is less than 0! Analyzing this value, it can be concluded that the reaction will proceed in the forward direction and reactants will be depleted and products will be formed. In addition, Q (68.89) is less than K (130) and one could conclude that the reaction will proceed in the forward direction from this comparison alone. As a result, this provides further evidence of the reaction proceeding forward and concurs with the ΔG value.
References
(1) The Science of Fake Blood. https://www.insidescience.org/news/science-fake-blood (accessed Dec 4, 2022).
(2) Physical Chemistry | Chemical equilibrium. https://www.ld-didactic.de/documents...C4/C4211_e.pdf (accessed Dec 7, 2022).
(3) Chemical Equilibrium: Determination of an Equilibrium Constant of a Complex. https://staff.buffalostate.edu/nazar...%20that,SCN%2D)%20are%20practically%20colorless (accessed Nov 11, 2022).
(4) Free energy | endergonic vs Exergonic reactions (article). https://www.khanacademy.org/science/...bs-free-energy (accessed Nov 11, 2022).
Figures
(1) Motamayor, R. Polygon. Vox Media March 15, 2022. https://www.polygon.com/22979890/x-r...t-horror-movie (accessed Dec 7, 2022).