b) Draw an energy profile diagram showing energy of activation for endothermic reaction. [2] c) The hypothetical data for the reaction, 2A + B C , are, Expt No. [A] mol\,L^ 1 [B] mol\,L^ 1 [rate]…

b) Energy Profile Diagram for an Endothermic Reaction

The energy profile diagram for an endothermic reaction shows the energy changes during the reaction. The reactants have lower energy than the products, and the activation energy ($E_a$) is the energy barrier that must be overcome for the reaction to proceed.

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c) Rate Law and Calculation for the Reaction $2\mathrm{A}+\mathrm{B}\to \mathrm{C}$

Step 1: Determine the Rate Law

The rate law for the reaction is given by: $$\text{Rate}=k[A{]}^m[B{]}^n$$ where $k$ is the rate constant, and $m$ and $n$ are the orders of the reaction with respect to $A$ and $B$, respectively.

We use the given experimental data to find $m$ and $n$:

Step 2: Find the Order with Respect to $A$ ($m$)

Compare Expt 1 and Expt 2 (where $[B]$ changes but $[A]$ changes by a factor of 3): $$\frac{{\text{Rate}}_2}{{\text{Rate}}_1}=\frac{7.2\times 10^{-2}}{6\times 10^{-3}}=12$$ $$\frac{[A{]}_2}{[A{]}_1}=\frac{0.3}{0.1}=3$$ $$\frac{[B{]}_2}{[B{]}_1}=\frac{0.2}{0.1}=2$$ Since $[B]$ changes, we must isolate the effect of $[A]$ by comparing experiments where $[B]$ is constant. Compare Expt 1 and Expt 4 (where $[B]$ is constant at 0.1 mol/L): $$\frac{{\text{Rate}}_4}{{\text{Rate}}_1}=\frac{2.40\times 10^{-2}}{6\times 10^{-3}}=4$$ $$\frac{[A{]}_4}{[A{]}_1}=\frac{0.4}{0.1}=4$$ Since the rate increases by a factor of 4 when $[A]$ increases by a factor of 4, the order with respect to $A$ is: $$m=1$$

Step 3: Find the Order with Respect to $B$ ($n$)

Compare Expt 2 and Expt 3 (where $[A]$ is constant at 0.3 mol/L): $$\frac{{\text{Rate}}_3}{{\text{Rate}}_2}=\frac{2.88\times 10^{-1}}{7.2\times 10^{-2}}=4$$ $$\frac{[B{]}_3}{[B{]}_2}=\frac{0.4}{0.2}=2$$ Since the rate increases by a factor of 4 when $[B]$ doubles, the order with respect to $B$ is: $$n=2$$

Thus, the rate law is: $$\text{Rate}=k[A{]}^1[B{]}^2$$

Step 4: Calculate the Rate Constant ($k$)

Using Expt 1: $$6\times 10^{-3}=k(0.1{)}^1(0.1{)}^2$$ $$6\times 10^{-3}=k(0.1)(0.01)$$ $$6\times 10^{-3}=k(0.001)$$ $$k=\frac{6\times 10^{-3}}{0.001}=6{\text{L}}^2\cdot {\text{mol}}^{-2}\cdot {\text{s}}^{-1}$$

Step 5: Calculate the Rate for $[A]=0.6\text{mol/L}$ and $[B]=0.3\text{mol/L}$

Using the rate law: $$\text{Rate}=k[A][B{]}^2$$ $$\text{Rate}=6\times (0.6)\times (0.3{)}^2$$ $$\text{Rate}=6\times 0.6\times 0.09$$ $$\text{Rate}=6\times 0.054$$ $$\text{Rate}=0.324\text{mol/L⋅s}$$

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OR

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a) Deduce the Normality Equation

Normality ($N$) is defined as the number of gram equivalents of solute per liter of solution. The normality equation is derived as follows:

$$\text{Normality}=\frac{\text{Number of gram equivalents}}{\text{Volume of solution (in liters)}}$$

For an acid or base, the number of gram equivalents is given by: $$\text{Gram equivalents}=\frac{\text{Mass (g)}}{\text{Equivalent weight (g/equiv)}}$$

The equivalent weight is calculated as: $$\text{Equivalent weight}=\frac{\text{Molar mass}}{\text{Number of replaceable H⁺ or OH⁻ ions}}$$

For a redox reaction, the equivalent weight is: $$\text{Equivalent weight}=\frac{\text{Molar mass}}{\text{Change in oxidation number per molecule}}$$

Thus, the normality equation is: $$N=\frac{\text{Mass (g)}}{\text{Equivalent weight (g/equiv)}\times \text{Volume (L)}}$$

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b) Reasons

i) Methyl Orange as Indicator for $\mathrm{H}{}_2\mathrm{SO}{}_4$ and $\mathrm{Na}{}_2\mathrm{CO}{}_3$ Titration

Methyl orange is used because:

• The reaction between $\mathrm{H}{}_2\mathrm{SO}{}_4$ (a strong acid) and $\mathrm{Na}{}_2\mathrm{CO}{}_3$ (a weak base) produces $\mathrm{CO}{}_2$, water, and $\mathrm{Na}{}_2\mathrm{SO}{}_4$.
• The equivalence point occurs in the acidic range (pH ~ 4), where methyl orange changes color from red to yellow.

ii) Warming Oxalic Acid Solution with Dilute $\mathrm{H}{}_2\mathrm{SO}{}_4$ Before Titration with $\mathrm{KMnO}{}_4$

Oxalic acid ($\mathrm{H}{}_2\mathrm{C}{}_2\mathrm{O}{}_4$) is warmed with dilute $\mathrm{H}{}_2\mathrm{SO}{}_4$ to:

• Remove dissolved $\mathrm{CO}{}_2$ gas, which can interfere with the titration.
• Ensure complete dissociation of oxalic acid into oxalate ions ($\mathrm{C}{}_2\mathrm{O}{}_4{}^{2-}$), which react with $\mathrm{KMnO}{}_4$ in acidic medium.

Oxalic acidOxalic acid

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c) Calculation for $\mathrm{Na}{}_2\mathrm{CO}{}_3\cdot x\mathrm{H}{}_2\mathrm{O}$ Sample

Given:

• Mass of $\mathrm{Na}{}_2\mathrm{CO}{}_3\cdot x\mathrm{H}{}_2\mathrm{O}$ = 0.70 g
• Volume made = 100 mL
• 20 mL of this solution requires 19.8 mL of $\mathrm{KMnO}{}_4$ solution for titration.

Step 1: Determine the Molarity of $\mathrm{KMnO}{}_4$ Solution

Assume the normality of $\mathrm{KMnO}{}_4$ is $N$ (given in the question, but since it is missing, we assume a standard value for demonstration). However, since the question is incomplete, we proceed with the general approach:

The reaction between oxalic acid and $\mathrm{KMnO}{}_4$ is: $$5\mathrm{C}{}_2\mathrm{O}{}_4{}^{2-}+2\mathrm{MnO}{}_4{}^{-}+16\mathrm{H}{}^{+}\to 10\mathrm{CO}{}_2+2\mathrm{Mn}{}^{2+}+8\mathrm{H}{}_2\mathrm{O}$$

The normality equation for $\mathrm{KMnO}{}_4$ is: $$N_{\mathrm{KMnO}{}_4}=\frac{\text{Number of equivalents of }\mathrm{KMnO}{}_4}{\text{Volume (L)}}$$

Since the exact normality of $\mathrm{KMnO}{}_4$ is not provided, we assume it is known (e.g., $N_{\mathrm{KMnO}{}_4}=0.02\text{N}$ for calculation purposes). The actual calculation would require the normality of $\mathrm{KMnO}{}_4$ from a standard solution.

Step 2: Calculate the Moles of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ in the Sample

The titration reaction for $\mathrm{Na}{}_2\mathrm{CO}{}_3$ with $\mathrm{KMnO}{}_4$ (assuming indirect titration via oxalic acid) is complex, but the key step is: $$\text{Equivalents of }\mathrm{Na}{}_2\mathrm{CO}{}_3=\text{Equivalents of }\mathrm{KMnO}{}_4$$

Given:

• Volume of $\mathrm{KMnO}{}_4$ used = 19.8 mL = 0.0198 L
• Normality of $\mathrm{KMnO}{}_4$ = $N$ (assumed known)

$$\text{Equivalents of }\mathrm{KMnO}{}_4=N\times 0.0198$$

Since 20 mL of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ solution corresponds to this volume of $\mathrm{KMnO}{}_4$, the normality of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ solution is: $$N_{\mathrm{Na}{}_2\mathrm{CO}{}_3}=\frac{N\times 0.0198}{0.020}$$

The mass of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ in 100 mL is then calculated, and the value of $x$ (water of crystallization) can be determined using the molar mass of $\mathrm{Na}{}_2\mathrm{CO}{}_3\cdot x\mathrm{H}{}_2\mathrm{O}$.

Since the exact normality of $\mathrm{KMnO}{}_4$ is missing, the final calculation cannot be completed here. However, the general approach is as follows:

1. Calculate the equivalents of $\mathrm{KMnO}{}_4$ used.
2. Relate it to the equivalents of $\mathrm{Na}{}_2\mathrm{CO}{}_3$ in the sample.
3. Use the molar mass of anhydrous $\mathrm{Na}{}_2\mathrm{CO}{}_3$ (106 g/mol) and the given mass (0.70 g) to find $x$.

For example, if $N_{\mathrm{KMnO}{}_4}=0.02\text{N}$: $$\text{Equivalents of }\mathrm{KMnO}{}_4=0.02\times 0.0198=0.000396$$ $$\text{Equivalents of }\mathrm{Na}{}_2\mathrm{CO}{}_3=0.000396$$ $$\text{Moles of }\mathrm{Na}{}_2\mathrm{CO}{}_3=\frac{0.000396}{2}=0.000198\text{mol}$$ $$\text{Mass of anhydrous }\mathrm{Na}{}_2\mathrm{CO}{}_3=0.000198\times 106=0.021\text{g}$$ $$\text{Mass of water}=0.70-0.021=0.679\text{g}$$ $$\text{Moles of water}=\frac{0.679}{18}=0.0377\text{mol}$$ $$x=\frac{0.0377}{0.000198}\approx 190$$

This value of $x$ is unrealistic, indicating that the assumed normality of $\mathrm{KMnO}{}_4$ is incorrect. The actual calculation requires the correct normality of $\mathrm{KMnO}{}_4$ from the question. The final answer cannot be determined without the normality of $\mathrm{KMnO}{}_4$.