ESAT Chemistry Mock 3

First, determine the composition of the layers. P and Q are miscible, so they form a single mixed phase. R is immiscible with both, so it forms a separate phase.

Next, determine the position of the layers using density. The density of the P + Q mixture will be intermediate between 0.79 and 1.00 (approx. $\displaystyle 0.90\,\text{g cm}^{-3}$ ), which is significantly lower than the density of R ( $\displaystyle 1.49\,\text{g cm}^{-3}$ ). Therefore, the upper layer contains the P + Q mixture, and the lower layer contains pure R.

Finally, to separate P from the P + Q mixture (upper layer), fractional distillation is used. P has the lower boiling point ( $\displaystyle 56^\circ\text{C}$ ) compared to Q ( $\displaystyle 100^\circ\text{C}$ ). The mixture should be heated to ** $\displaystyle 56^\circ\text{C}$ ** to vaporise and collect P.

The energy transferred ( $\displaystyle q$ ) is given by the equation:Rearranging for temperature change:In Experiment 1, the temperature rise is $\displaystyle \Delta T$ .

In Experiment 2, the energy $\displaystyle q$ is the same, but the mass is $\displaystyle 3m$ and the specific heat capacity is $\displaystyle \dfrac{1}{2}c$ . Substituting these values into the equation for the new temperature rise ( $\displaystyle \Delta T_2$ ):Since $\displaystyle \dfrac{q}{mc} = \Delta T$ , we can substitute this back in:

Let us evaluate the effect of each action on initial forward rate and equilibrium yield:

- Action 1 (decreasing the volume at constant temperature):
- Decreasing volume increases the concentrations of gaseous reactants, which increases the collision frequency and therefore increases the initial rate of the forward reaction for both reactions.
- For Reaction 1, there are 3 moles of gas on the reactant side and 1 mole of gas on the product side ( $\displaystyle 3 \to 1$ ). Decreasing volume increases total pressure, shifting the equilibrium to the side with fewer moles of gas (forward), thereby **increasing the equilibrium yield of $\displaystyle \ce{CH3OH(g)}$ **.
- For Reaction 2, there are 2 moles of gas on the reactant side and 2 moles of gas on the product side ( $\displaystyle 2 \to 2$ ). Because $\displaystyle \Delta n_{\text{gas}} = 0$ , changes in pressure/volume have no effect on the equilibrium position or yield.
- Therefore, Action 1 satisfies all criteria.

- Action 2 (increasing the temperature at constant volume):
- Increases the initial rate for both reactions.
- Reaction 1 is exothermic ( $\displaystyle \Delta H < 0$ ). By Le Chatelier's principle, increasing temperature shifts the equilibrium in the endothermic (reverse) direction, decreasing the equilibrium yield of $\displaystyle \ce{CH3OH}$ .
- Reaction 2 is endothermic ( $\displaystyle \Delta H > 0$ ), so increasing temperature shifts equilibrium to the right, increasing its yield (rather than having no effect).
- Therefore, Action 2 does not satisfy the criteria.

- Action 3 (adding an appropriate catalyst):
- Increases the initial forward rate for both reactions by providing an alternative pathway with lower activation energy.
- Catalysts speed up both the forward and reverse reactions equally, so a catalyst has no effect on the equilibrium yield of products for either reaction. Thus, it fails to increase the equilibrium yield for Reaction 1.

Hence, only action 1 satisfies all conditions.

Statements 1 and 3 are correct. Greenhouse warming is associated with reduced loss of infrared radiation, not mainly reflection of visible light, so statement 2 is incorrect.

To balance the equation, express the coefficients in terms of $\displaystyle a$ by applying the conservation of atoms for each element:

1. Carbon (C): $\displaystyle 3a = c \implies c = 3a$ 2. Hydrogen (H): $\displaystyle 7a = 2d \implies d = \dfrac{7}{2}a$ 3. Nitrogen (N): $\displaystyle a = 2e \implies e = \dfrac{1}{2}a$ 4. Sulfur (S): $\displaystyle a = f \implies f = a$ 5. Oxygen (O): $\displaystyle \text{Total O on left} = 2a + 2b$ $\displaystyle \text{Total O on right} = 2c + d + 2f = 2(3a) + \dfrac{7}{2}a + 2(a) = \dfrac{23}{2}a$ Equating oxygen atoms on both sides: $\displaystyle 2a + 2b = \dfrac{23}{2}a$ $\displaystyle 2b = \dfrac{19}{2}a \implies b = \dfrac{19}{4}a$ To obtain the smallest whole-number coefficients, set $\displaystyle a = 4$ :
- $\displaystyle a = 4$ - $\displaystyle b = 19$ - $\displaystyle c = 12$ - $\displaystyle d = 14$ - $\displaystyle e = 2$ - $\displaystyle f = 4$ Checking the balanced equation: $\displaystyle \ce{4 C3H7NO2S + 19 O2 -> 12 CO2 + 14 H2O + 2 N2 + 4 SO2}$ Thus, the coefficient $\displaystyle b = 19$ .

To find the balanced chemical equation, determine whether each species is a reactant or product, and calculate its change in concentration ( $\displaystyle \Delta [\text{concentration}]$ ):

1. Reactants (concentrations decrease):
- $\displaystyle \ce{W}$ : decreases from $\displaystyle 1.20\text{ mol dm}^{-3}$ to $\displaystyle 0.00\text{ mol dm}^{-3}$ , so $\displaystyle \Delta[\ce{W}] = -1.20\text{ mol dm}^{-3}$ .
- $\displaystyle \ce{X}$ : decreases from $\displaystyle 0.80\text{ mol dm}^{-3}$ to $\displaystyle 0.40\text{ mol dm}^{-3}$ , so $\displaystyle \Delta[\ce{X}] = -0.40\text{ mol dm}^{-3}$ .

2. Products (concentrations increase):
- $\displaystyle \ce{Y}$ : increases from $\displaystyle 0.00\text{ mol dm}^{-3}$ to $\displaystyle 0.80\text{ mol dm}^{-3}$ , so $\displaystyle \Delta[\ce{Y}] = +0.80\text{ mol dm}^{-3}$ .
- $\displaystyle \ce{Z}$ : increases from $\displaystyle 0.20\text{ mol dm}^{-3}$ to $\displaystyle 0.60\text{ mol dm}^{-3}$ , so $\displaystyle \Delta[\ce{Z}] = +0.40\text{ mol dm}^{-3}$ .

3. Stoichiometric ratio:
The ratio of changes in concentration is: $\displaystyle -\Delta[\ce{W}] : -\Delta[\ce{X}] : \Delta[\ce{Y}] : \Delta[\ce{Z}] = 1.20 : 0.40 : 0.80 : 0.40 = 3 : 1 : 2 : 1$ Therefore, the balanced equation is $\displaystyle \ce{3W + X -> 2Y + Z}$ .

During the electrolysis of dilute $\displaystyle \ce{H2SO4(aq)}$ , the electrode reactions are:

Cathode (negative electrode, reduction): $\displaystyle \ce{2H+(aq) + 2e- -> H2(g)}$ Anode (positive electrode, oxidation): $\displaystyle \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}$ The overall equation is $\displaystyle \ce{2H2O(l) -> 2H2(g) + O2(g)}$ , producing a $\displaystyle 2:1$ molar (and volume) ratio of $\displaystyle \ce{H2}$ to $\displaystyle \ce{O2}$ .

From the graph, electrode X produces twice the volume of gas compared to electrode Y ( $\displaystyle 80\text{ cm}^3$ vs $\displaystyle 40\text{ cm}^3$ at $\displaystyle 20\text{ min}$ ). Therefore, gas X is $\displaystyle \ce{H2}$ and gas Y is $\displaystyle \ce{O2}$ .

- Statement 1 is correct: $\displaystyle \ce{H2}$ is produced at the cathode, which is connected to the negative terminal of the d.c. power supply.
- Statement 2 is incorrect: Oxidation releases electrons at the anode (electrode Y), and reduction consumes electrons at the cathode (electrode X). Thus, in the external circuit, electrons flow from electrode Y to electrode X.
- Statement 3 is correct: The half-equation at the anode shows that producing $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{O2}$ (gas Y) requires the transfer of $\displaystyle 4\text{ mol}$ of electrons through the external circuit.

Therefore, statements 1 and 3 only are correct (option F).

1. Elements W and Y displace bromine from $\displaystyle \ce{KBr(aq)}$ , meaning they are stronger oxidising agents (more reactive) than bromine. Elements X and Z cannot displace bromine from $\displaystyle \ce{KBr(aq)}$ , meaning they are weaker oxidising agents (less reactive) than bromine. Therefore, both W and Y are more reactive than both X and Z: $\displaystyle \{\text{W}, \text{Y}\} > \{\text{X}, \text{Z}\}$ .

2. Element Y reacts explosively with $\displaystyle \ce{H2}$ in the dark, whereas element W requires ultraviolet light. Reactivity of non-metals with hydrogen decreases as oxidising power decreases. Therefore, Y is more reactive than W: $\displaystyle \text{Y} > \text{W}$ .

3. Element Z displaces element X from an aqueous solution of $\displaystyle \ce{NaX}$ . A more reactive non-metal displaces a less reactive one from its salt solution, so Z is more reactive than X: $\displaystyle \text{Z} > \text{X}$ .

Combining these results gives the order from most reactive to least reactive: $\displaystyle \text{Y} > \text{W} > \text{Z} > \text{X}$ Hence, the correct option is F.

First, determine the group of element X from its atomic number ( $\displaystyle Z=15$ ).

The electron configuration is $\displaystyle 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3$ .

The outer shell ( $\displaystyle n=3$ ) has $\displaystyle 2 + 3 = 5$ valence electrons, placing X in Group 15.

To complete its octet, X needs 3 electrons, so it forms 3 single bonds with hydrogen atoms. The formula is $\displaystyle \text{XH}_3$ .

Next, determine the shape using VSEPR theory:

- Valence electrons on X: 5

- Electrons used in bonding: 3 (one for each H)

- Remaining electrons: $\displaystyle 5 - 3 = 2$ (1 lone pair)

There are 3 bonding pairs and 1 lone pair (4 electron domains in total). The electron arrangement is tetrahedral, but the molecular geometry is determined by the atom positions only, resulting in a trigonal pyramidal shape.

1. Baseline Experiment (Curve X):
- Amount of $\displaystyle \ce{CaCO3} = \dfrac{2.00\text{ g}}{100\text{ g mol}^{-1}} = 0.020\text{ mol}$ .
- Amount of $\displaystyle \ce{HCl} = 0.050\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.050\text{ mol}$ .
- Stoichiometry requires $\displaystyle 2\text{ mol } \ce{HCl}$ per $\displaystyle 1\text{ mol } \ce{CaCO3}$ , so $\displaystyle 0.020\text{ mol } \ce{CaCO3}$ requires $\displaystyle 0.040\text{ mol } \ce{HCl}$ .
- $\displaystyle \ce{CaCO3}$ is the limiting reagent, producing $\displaystyle 0.020\text{ mol}$ of $\displaystyle \ce{CO2}$ (represented by a final height of 4 grid units).

2. Experiment 1:
- Amount of $\displaystyle \ce{CaCO3} = \dfrac{3.00\text{ g}}{100\text{ g mol}^{-1}} = 0.030\text{ mol}$ .
- Amount of $\displaystyle \ce{HCl} = 0.050\text{ mol}$ .
- To react all $\displaystyle 0.030\text{ mol } \ce{CaCO3}$ would require $\displaystyle 0.060\text{ mol } \ce{HCl}$ . Therefore, $\displaystyle \ce{HCl}$ is now the limiting reagent.
- Amount of $\displaystyle \ce{CO2}$ formed $\displaystyle = \dfrac{0.050}{2} = 0.025\text{ mol}$ , which is $\displaystyle \dfrac{0.025}{0.020} = 1.25\times$ the baseline volume ( $\displaystyle 1.25 \times 4 = 5$ grid units).
- Powdered $\displaystyle \ce{CaCO3}$ provides a larger surface area, giving a greater initial rate (steeper initial gradient).
- This corresponds to curve 2.

3. Experiment 2:
- Amount of $\displaystyle \ce{CaCO3} = 0.020\text{ mol}$ .
- Amount of $\displaystyle \ce{HCl} = 0.100\text{ dm}^3 \times 0.40\text{ mol dm}^{-3} = 0.040\text{ mol}$ .
- The reagents are in exact stoichiometric ratio ( $\displaystyle 0.040 / 2 = 0.020\text{ mol}$ ), producing $\displaystyle 0.020\text{ mol}$ of $\displaystyle \ce{CO2}$ (the same final volume as the baseline, 4 grid units).
- The lower concentration of $\displaystyle \ce{HCl}$ ( $\displaystyle 0.40\text{ mol dm}^{-3}$ vs $\displaystyle 1.0\text{ mol dm}^{-3}$ ) results in a lower initial rate (shallower initial gradient).
- This corresponds to curve 4.

Therefore, the correct row is C.

Set the enthalpy of the initial reactant $\displaystyle \ce{P}$ as the reference level $\displaystyle H(\ce{P}) = 0\text{ kJ mol}^{-1}$ .

1. **First transition state ( $\displaystyle \text{TS}_1$ ):**
The forward activation energy for $\displaystyle \ce{P -> Q}$ is $\displaystyle 58\text{ kJ mol}^{-1}$ , so: $\displaystyle H(\text{TS}_1) = 0 + 58 = +58\text{ kJ mol}^{-1}$ 2. **Intermediate $\displaystyle \ce{Q}$ :** $\displaystyle H(\ce{Q}) = +22\text{ kJ mol}^{-1}$ Therefore, the activation energy for the reverse step $\displaystyle \ce{Q -> P}$ is: $\displaystyle E_{\text{a}}(\ce{Q -> P}) = H(\text{TS}_1) - H(\ce{Q}) = 58 - 22 = 36\text{ kJ mol}^{-1}$ 3. **Second transition state ( $\displaystyle \text{TS}_2$ ):**
The activation energy for $\displaystyle \ce{Q -> R}$ is $\displaystyle 74\text{ kJ mol}^{-1}$ relative to $\displaystyle \ce{Q}$ : $\displaystyle H(\text{TS}_2) = H(\ce{Q}) + 74 = 22 + 74 = +96\text{ kJ mol}^{-1}$ 4. **Product $\displaystyle \ce{R}$ :**
The overall enthalpy change is $\displaystyle \Delta H = -35\text{ kJ mol}^{-1}$ , so: $\displaystyle H(\ce{R}) = -35\text{ kJ mol}^{-1}$ Therefore, the activation energy for the conversion $\displaystyle \ce{R -> Q}$ is the barrier from $\displaystyle \ce{R}$ to $\displaystyle \text{TS}_2$ : $\displaystyle E_{\text{a}}(\ce{R -> Q}) = H(\text{TS}_2) - H(\ce{R}) = 96 - (-35) = 131\text{ kJ mol}^{-1}$ Thus, $\displaystyle E_{\text{a}}(\ce{Q -> P}) = 36\text{ kJ mol}^{-1}$ and $\displaystyle E_{\text{a}}(\ce{R -> Q}) = 131\text{ kJ mol}^{-1}$ , corresponding to option C.

During a change of state, energy is used to overcome attractive forces and change particle arrangement rather than raise the temperature.

1. Calculate the mass of limonene: $\displaystyle \text{mass} = \text{volume} \times \text{density} = 1.60\text{ cm}^3 \times 0.85\text{ g cm}^{-3} = 1.36\text{ g}$ 2. Calculate the amount in moles of limonene: $\displaystyle n(\text{limonene}) = \dfrac{1.36\text{ g}}{136\text{ g mol}^{-1}} = 0.010\text{ mol}$ 3. Determine the stoichiometry:
From the skeletal structure, limonene contains $\displaystyle 2$ $\displaystyle \ce{C=C}$ double bonds (one in the six-membered ring and one in the isopropenyl group). Therefore, complete hydrogenation requires $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{H2}$ per mole of limonene: $\displaystyle n(\ce{H2}) = 2 \times 0.010\text{ mol} = 0.020\text{ mol}$ 4. Calculate the volume of $\displaystyle \ce{H2(g)}$ at RTP: $\displaystyle V(\ce{H2}) = 0.020\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.48\text{ dm}^3$

1. Determine the molecular formula and relative molecular mass ( $\displaystyle M_\text{r}$ ) of the compound (2-hydroxypropanoic acid): $\displaystyle \text{Formula} = \ce{C3H6O3}$ $\displaystyle M_\text{r} = (3 \times 12.0) + (6 \times 1.0) + (3 \times 16.0) = 36.0 + 6.0 + 48.0 = 90.0$ 2. Calculate the moles of the compound in $\displaystyle 0.45\text{ g}$ : $\displaystyle n = \dfrac{0.45\text{ g}}{90.0\text{ g mol}^{-1}} = 0.0050\text{ mol}$ 3. Determine the stoichiometry of the reaction with sodium metal:
Sodium reacts with both the alcohol group ( $\displaystyle -\ce{OH}$ ) and the carboxylic acid group ( $\displaystyle -\ce{COOH}$ ): $\displaystyle \ce{-OH + Na -> -ONa + 1/2 H2}$ $\displaystyle \ce{-COOH + Na -> -COONa + 1/2 H2}$ Each molecule of 2-hydroxypropanoic acid contains one $\displaystyle -\ce{OH}$ group and one $\displaystyle -\ce{COOH}$ group, so each mole of compound produces: $\displaystyle \dfrac{1}{2} + \dfrac{1}{2} = 1.0\text{ mol of } \ce{H2(g)}$ $\displaystyle \ce{CH3CH(OH)COOH + 2Na -> CH3CH(ONa)COONa + H2(g)}$ 4. Calculate the volume of $\displaystyle \ce{H2}$ gas produced: $\displaystyle n(\ce{H2}) = 0.0050\text{ mol}$ $\displaystyle V = 0.0050\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.120\text{ dm}^3$ $\displaystyle V = 0.120 \times 1000\text{ cm}^3 = 120\text{ cm}^3$ Thus, the correct option is C.

A green precipitate with $\displaystyle \ce{NaOH}$ identifies $\displaystyle \ce{Fe^{2+}}$ , while the acidified barium test identifies sulfate, giving $\displaystyle \ce{FeSO4}$ .

All three are specified common properties of transition metals.

1. Frequency of collisions: Grinding the solid calcium carbonate into a fine powder significantly increases the exposed surface area of the solid. With more $\displaystyle \ce{CaCO3}$ particles accessible to the aqueous $\displaystyle \ce{H+}$ ions at any given instant, the frequency of collisions between reactant particles increases.

2. Activation energy: Activation energy is the minimum kinetic energy that colliding particles must possess for a reaction to occur. It depends exclusively on the chemical reaction pathway. Changing the physical surface area of a reactant does not alter the reaction mechanism or pathway, so the activation energy remains unchanged (only a catalyst changes the activation energy).

3. Initial rate of reaction: Because the frequency of collisions increases while the proportion of successful collisions (those with energy $\displaystyle E \ge E_\text{a}$ ) remains unchanged at constant temperature, the frequency of successful collisions per unit time increases, and therefore the initial rate of reaction increases.

Hence, row G is correct.

We must identify the constituent elements of each solute and compare them to the products formed at the cathode and anode during concentrated aqueous electrolysis.

1. ** $\displaystyle \ce{NaCl}$ (Constituent elements: Na, Cl)
- Cathode:** Hydrogen ( $\displaystyle \ce{H2}$ ) is produced because sodium is more reactive than hydrogen.
- Anode: Chlorine ( $\displaystyle \ce{Cl2}$ ) is produced (concentrated halide).
- Result: Products are H and Cl. Na is missing. (Incorrect)

2. ** $\displaystyle \ce{HBr}$ (Constituent elements: H, Br)
- Cathode:** Hydrogen ( $\displaystyle \ce{H2}$ ) is produced.
- Anode: Bromine ( $\displaystyle \ce{Br2}$ ) is produced (concentrated halide).
- Result: Products are H and Br. Both constituent elements are present. (Correct)

3. ** $\displaystyle \ce{CuCl2}$ (Constituent elements: Cu, Cl)
- Cathode:** Copper ( $\displaystyle \ce{Cu}$ ) is produced because copper is less reactive than hydrogen.
- Anode: Chlorine ( $\displaystyle \ce{Cl2}$ ) is produced (concentrated halide).
- Result: Products are Cu and Cl. Both constituent elements are present. (Correct)

Therefore, solutions 2 and 3 meet the condition.

First, determine the initial oxidation states of the central atoms:

* In $\displaystyle \ce{[PtCl4]^{2-}}$ , chloride is $\displaystyle -1$ . The oxidation state of Pt is calculated as: $\displaystyle x + 4(-1) = -2 \implies x = +2$ .
* In $\displaystyle \ce{MnO4^{-}}$ , oxygen is $\displaystyle -2$ . The oxidation state of Mn is calculated as: $\displaystyle y + 4(-2) = -1 \implies y = +7$ .

Next, calculate the change in oxidation number (electrons transferred) per ion:

* Pt oxidises from $\displaystyle +2$ to $\displaystyle +4$ , a loss of $\displaystyle 2$ electrons per Pt atom.
* Mn reduces from $\displaystyle +7$ to $\displaystyle +2$ , a gain of $\displaystyle 5$ electrons per Mn atom.

To balance the electrons, the total electrons lost must equal the total electrons gained. The lowest common multiple of $\displaystyle 2$ and $\displaystyle 5$ is $\displaystyle 10$ .Therefore, we need $\displaystyle 5$ moles of $\displaystyle \ce{[PtCl4]^{2-}}$ for every $\displaystyle 2$ moles of $\displaystyle \ce{MnO4^{-}}$ .

The ratio is $\displaystyle 5 : 2$ .

1. Find the mass and moles of oxygen combined with metal $\displaystyle \ce{M}$ : $\displaystyle \text{Mass of } \ce{O} = 6.552\text{ g} - 4.872\text{ g} = 1.680\text{ g}$ $\displaystyle n(\ce{O}) = \dfrac{1.680\text{ g}}{16.0\text{ g mol}^{-1}} = 0.105\text{ mol}$ 2. Use the stoichiometric ratio from the formula $\displaystyle \ce{M2O3}$ to find the moles of $\displaystyle \ce{M}$ : $\displaystyle \dfrac{n(\ce{M})}{n(\ce{O})} = \dfrac{2}{3} \implies n(\ce{M}) = \dfrac{2}{3} \times 0.105\text{ mol} = 0.070\text{ mol}$ 3. Calculate the relative atomic mass $\displaystyle A_{\text{r}}(\ce{M})$ : $\displaystyle A_{\text{r}}(\ce{M}) = \dfrac{4.872\text{ g}}{0.070\text{ mol}} = 69.60$ 4. Let $\displaystyle x$ be the percentage abundance of $\displaystyle ^{69}\ce{M}$ . The abundance of $\displaystyle ^{71}\ce{M}$ is $\displaystyle (100 - x)\%$ : $\displaystyle \dfrac{69.0 x + 71.0(100 - x)}{100} = 69.60$ $\displaystyle 71.0 - 0.02 x = 69.60$ $\displaystyle 0.02 x = 1.40 \implies x = 70\%$

Since $\displaystyle 112\text{ kJ}$ of energy is released per mole of $\displaystyle \ce{CO}$ reacted, the reaction is exothermic, so $\displaystyle \Delta H = -112\text{ kJ mol}^{-1}$ .

The enthalpy change is given by: $\displaystyle \Delta H = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$ Bonds broken (reactants):
- $\displaystyle 1 \times \ce{C-O}$ bond in $\displaystyle \ce{CO}$ (let this be $\displaystyle E_{\ce{CO}}$ )
- $\displaystyle 2 \times (\ce{H-H}) = 2 \times 436 = 872\text{ kJ mol}^{-1}$ $\displaystyle \text{Total broken} = E_{\ce{CO}} + 872\text{ kJ mol}^{-1}$ Bonds formed (products):
In $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{CH3OH}$ :
- $\displaystyle 3 \times (\ce{C-H}) = 3 \times 413 = 1239\text{ kJ mol}^{-1}$ - $\displaystyle 1 \times (\ce{C-O}) = 358\text{ kJ mol}^{-1}$ - $\displaystyle 1 \times (\ce{O-H}) = 464\text{ kJ mol}^{-1}$ $\displaystyle \text{Total formed} = 1239 + 358 + 464 = 2061\text{ kJ mol}^{-1}$ **Calculating $\displaystyle E_{\ce{CO}}$ :** $\displaystyle \Delta H = (E_{\ce{CO}} + 872) - 2061 = -112$ $\displaystyle E_{\ce{CO}} - 1189 = -112$ $\displaystyle E_{\ce{CO}} = 1189 - 112 = 1077\text{ kJ mol}^{-1}$