# Tutor profile: Angelo C.

## Questions

### Subject: Chemistry

Carbon dioxide gas, $$CO_2(g)$$, is produced from the reaction of carbon monoxide, $$CO(g)$$, and water, $$H_2O$$, according to the following reaction: \begin{equation} CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \end{equation} Starting with 1 mol each of $$CO_2(g)$$ and $$H_2O$$ in a 50 L vessel. How many mols of each substance are in this mixture of equilibrium at 1000$$^{\circ}$$C where the equilibrium constant is 0.58?

For this reaction, the equilibrium constant, $$K_C$$ can be expressed as follows: \begin{equation} K_C = \frac{ [CO_2(g)] [H_2(g)] } { [CO(g)] [H_2O (g)] } \end{equation} Additionally, the concentration of $$CO_2 (g)$$ and $$H_2O(g)$$ can be found using the standard definition of molarity: \begin{gather} Molarity (M) = \frac{mol}{volume} \\ \\ \therefore [C0_2(g)] = [H_2O(g)] = \frac{1 mol}{50 L} = 0.02 M \end{gather} At equilibrium, the products $$CO_2(g)$$ and $$H_2(g)$$ will be generated to a concentration $$x$$, and the reactants will each be depleted by a concentration $$x$$. Plugging this into our equation for the equilibrium constant yields the following: \begin{equation} 0.58 = \frac{(x)(x)} {(0.02-x)(0.02-x)} = \frac{x^2}{(0.02-x)^2} = \left( \frac{x}{0.02-x} \right)^2 \end{equation} Taking the square root of both sides -- and keeping more significant figures than required -- yields the following: \begin{align*} \pm 0.761577311 &= \frac{x}{0.02-x} \\ \end{align*} And now we evaluate the positive and negative cases separately. Positive Case: \begin{align*} 0.761577311 &= \frac{x}{0.02-x} \\ 0.01523 - 0.76158x &= x \\ x &= 8.65x10^-3 M \end{align*} Negative Case: \begin{align*} -0.761577311 &= \frac{x}{0.02-x} \\ -0.01523 + 0.76158x &= x \\ x &= -15.653 \Rightarrow \text{Reject the negative value of } x \end{align*} Now knowing the value of $$x$$ we can solve for the equilibrium concentrations of all our reagents and products: \begin{equation} \boxed { [CO_2(g)]_{eq} = [H_2(g)]_{eq} = 8.65x10^{-3} M \\ \\ [CO(g)]_{eq} = [H_2O]_{eq} = 0.02 - 8.65x10^{-3} M = 1.135x10^{-2} M } \end{equation}

### Subject: Biochemistry

For a Michaelis-Menton enzyme reaction $$ k_1 = 5x10^7 M^{-1} s^{-1} $$ , $$ k_{-1} = 2x10^4 s^{-1} $$ , and $$ k_2 = 4x10^2 s^{-1} $$. Calculate $$K_S$$ and $$K_M$$ for this reaction. Does substrate binding achieve equilibrium or steady state?

\begin{equation} K_M = \frac{ k_{-1} + k_2 }{k_1} = \frac{ \left( 2x10^4 s^{-1} \right) \left( 4x10^2 s^{-1} \right) } { 5x10^7 M^{-1}s^{-1}} = \boxed{4.08x10^{-4} M} \end{equation} With our value for $$K_M$$, we can now find $$K_S$$: \begin{equation} K_S = K_M - \frac{k_2}{k_1} = 4.08x10^{-4} M - \frac{4x10^2 s^{-1} } {5x10^7 M^{-1} s^{-1} } = \boxed{4.0x10^{-4} M} \end{equation} In this case, substrate binding achieves equilibrium since $$k_{-1} \gg k_2 $$. Given the values of $$k_{-1}$$ and $$k_2$$, the substrate is 100 times more likely to disassociate from the enzyme-substrate complex then go on to form the product.

### Subject: Chemical Engineering

You have a reaction scheme as follows: $$A \xrightarrow{\mathit{k_a}} I \xrightarrow{\mathit{k_I}} P $$ At time $$t=0$$ only reactant $$A$$ is present. What is the time dependence of $$[P]$$ using the steady state approximation?

Using the principles of mass action kinetics, you can write out the following differential equations for the concentrations f $$A$$, $$I$$, and $$P$$ as a function of time: \begin{equation} \frac{d[A]}{dt} = -k_A[A] \tag{1} \end{equation} \begin{equation} \frac{d[I]}{dt} = k_A[A] - k_I[I] \tag{2} \end{equation} \begin{equation} \frac{d[P]}{dt} = k_I[I] \tag{3} \end{equation} According to the steady-state assumption, the concentration of the intermediate $$I$$ does not change over time. Therefore we can write the following: \begin{equation} \frac{d[I]}{dt} = k_A[A] - k_I[I] = 0 \quad \Rightarrow \quad k_A[A] = k_I[I] \tag{4} \end{equation} Now we can rearrange and integrate equation $$(1)$$: \begin{equation} \int{ \frac{d[A]}{[A]} } = -k_A\int{ dt } \tag{5} \end{equation} \begin{equation} \therefore \ln{[A]} = -k_{A}t+C, \quad \text{where } C \text{ represents a constant} \tag{6} \end{equation} \begin{equation} [A] = Ce^{-k_At}, \tag{7} \end{equation} but since at $$t=0$$, $$[A] = [A]_0$$, we are left with the following as the solution: \begin{equation} [A] = [A]_0e^{-k_At} \tag{8} \end{equation} Now that we have an expression for $$[A]$$, the concentration of species $$A$$ overtime, we can substitute this into equation $$(4)$$ to get the following: \begin{equation} \frac{k_A}{k_I}[A]_0 e^{-k_At} = [I] \tag{9} \end{equation} Now we can substitute this expression for $$[I]$$ in equation $$(3)$$: \begin{align*} \frac{d[P]}{dt} &= k_I[I] \tag{10a} \\ &= k_I \left( \frac{k_A}{k_I}[A]_0 e^{-k_At} \right) \tag{10b} \\ &= k_A[A]_0e^{-k_At} \tag{10c} \end{align*} Now we can integrate the equation $$(10c)$$ to arrive at an expression for $$[P]$$: \begin{align*} \int{ d[P] } &= k_A [A]_0 \int{ e^{-k_At} dt } \tag{11a} \\ \therefore [P] &= k_A [A]_0 \left[ \frac{1}{k_A} \left( 1 - e^{-k_At} \right) \right] \tag{11b} \\ \end{align*} Which gives us our final expression for $$[P]$$: \begin{equation} \boxed{ [P] = [A]_0 \left( 1 - e^{-k_At} \right) } \end{equation}

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