Question

1. Calculate the standard enthalpy change for the reaction:
C2H4(g) + H2(g)→C2H6(g)
given that the enthalpy of combustion
for the reactants and products are:
∆Hºc(C2H4)(g) = -1411 kJ mol^-1∆Hºc(C2H6)(g) = -1560 kJ mol^-1∆Hºc(H2)(g) = -286 kJ mol-1.

2. Calculate the enthalpy change of
combustion for ethene gas (C2H4) given the
following enthalpy changes of formation:
∆Hºf(C2H4)(g) = +52 kJ mol^-1∆Hºf(CO2)(g) = -394 kJ mol^-1
∆Hºf(H2O)(g) = -286 kJ mol^-1

3.Calculate the standard entr
opy change for the reaction:
N2+ 3H2→2NH3
given the standard entropies
Sº(N2)(g) = 191.6 J K^-1 mol^-1
Sº(H2)(g) = 130.6 J K^-1 mol^-1
Sº(NH3)(g) = 193.3 J K^-1 mol^-1

Answer 1

Answer on Question #44400 - Chemistry - Other

Question:

1. Calculate the standard enthalpy change for the reaction:

$$\mathrm{C}_2\mathrm{H}_{4(g)} + \mathrm{H}_{2(g)} = \mathrm{C}_2\mathrm{H}_{6(g)}$$

given that the enthalpy of combustion, for the reactants and products are:

$$\Delta H_{c(\mathrm{C}_2\mathrm{H}_{4(g)})}^0 = -1411 \frac{kJ}{mol}; \Delta H_{c(\mathrm{C}_2\mathrm{H}_{6(g)})}^0 = -1560 \frac{kJ}{mol}; \Delta H_{c(\mathrm{H}_{2(g)})}^0 = -286 \frac{kJ}{mol}$$

Solution:

1. Write down the reactions of contribution of reagents and products:

$$\mathrm{C}_2\mathrm{H}_{4(g)} + 3\mathrm{O}_{2(g)} = 2\mathrm{CO}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(g)}, \Delta H_{c(\mathrm{C}_2\mathrm{H}_{4(g)})}^0 = -1411 \frac{kJ}{mol}; (I)$$ $$\mathrm{H}_{2(g)} + \frac{1}{2}\mathrm{O}_{2(g)} = \mathrm{H}_2\mathrm{O}_{(g)}, \Delta H_{c(\mathrm{H}_{2(g)})}^0 = -286 \frac{kJ}{mol}; (II)$$ $$\mathrm{C}_2\mathrm{H}_{6(g)} + \frac{7}{2}\mathrm{O}_{2(g)} = 2\mathrm{CO}_{2(g)} + 3\mathrm{H}_2\mathrm{O}_{(g)}, \Delta H_{c(\mathrm{C}_2\mathrm{H}_{6(g)})}^0 = -1560 \frac{kJ}{mol}; (III)$$

2. If we take the equation (III) with opposite sign and then add all three equations (I), (II) and (III), we will obtain the following:

$$\mathrm{C}_2\mathrm{H}_{4(g)} + 3\mathrm{O}_{2(g)} + \mathrm{H}_{2(g)} + \frac{1}{2}\mathrm{O}_{2(g)} + 2\mathrm{CO}_{2(g)} + 3\mathrm{H}_2\mathrm{O}_{(g)} = 2\mathrm{CO}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{H}_2\mathrm{O}_{(g)} + \mathrm{C}_2\mathrm{H}_{6(g)} + \frac{7}{2}\mathrm{O}_{2(g)}$$

3. The result is:

$$\mathrm{C}_2\mathrm{H}_{4(g)} + \mathrm{H}_{2(g)} = \mathrm{C}_2\mathrm{H}_{6(g)}$$

4. The standard enthalpy change for this reaction is equal to:

$$\Delta H^0 = \Sigma \Delta H_{c(\text{reagents})}^0 - \Sigma \Delta H_{c(\text{products})}^0 = -1411 \frac{kJ}{mol} + \left(-286 \frac{kJ}{mol}\right) - \left(-1560 \frac{kJ}{mol}\right) = -137 \frac{kJ}{mol}$$

Answer:

the standard enthalpy change for the reaction is $-137 \frac{kJ}{mol}$.

2. Calculate the enthalpy change of combustion for ethene gas $(\mathrm{C}_2\mathrm{H}_{4(g)})$ given the following enthalpy changes of formation:

$$\Delta H_{f(\mathrm{C}_2\mathrm{H}_{4(g)})}^0 = +52 \frac{kJ}{mol}; \Delta H_{f(\mathrm{CO}_{2(g)})}^0 = -394 \frac{kJ}{mol}; \Delta H_{f(\mathrm{H}_2\mathrm{O}_{(g)})}^0 = -286 \frac{kJ}{mol}$$

Solution:

1. Write down the reaction of contribution of ethane gas:

$$\mathrm{C_2H_4}(g) + 3\mathrm{O_2}(g) = 2\mathrm{CO_2}(g) + 2\mathrm{H_2O}(g), \Delta H^0$$

2. The standard enthalpy change for this reaction is equal to:

\begin{array}{l} \Delta H^0 = \Sigma \Delta H_f^0_{(products)} - \Sigma \Delta H_f^0_{(reagents)} \\ = 2 \cdot \Delta H_f^0_{(\mathrm{CO_2}(g))} + 2 \cdot \Delta H_f^0_{(\mathrm{H_2O}(g))} - \left(\Delta H_f^0_{(\mathrm{C_2H_4}(g))} + 3 \cdot \Delta H_f^0_{(\mathrm{O_2}(g))}\right) \end{array}

3. The enthalpy changes of formation of oxygen gas as an elementary substance equals zero:

\begin{array}{l} \Delta H^0 = 2 \cdot \Delta H_f^0_{(\mathrm{CO_2}(g))} + 2 \cdot \Delta H_f^0_{(\mathrm{H_2O}(g))} - \left(\Delta H_f^0_{(\mathrm{C_2H_4}(g))}\right) \\ = 2 \cdot \left(-394 \frac{kJ}{mol}\right) + 2 \cdot \left(-286 \frac{kJ}{mol}\right) - 52 \frac{kJ}{mol} = -1412 \frac{kJ}{mol} \end{array}

Answer: the standard enthalpy change for the reaction is $-1412 \frac{kJ}{mol}$.

3. Calculate the standard entropy change for the reaction:

$$\mathrm{N_2}(g) + 3\mathrm{H_2}(g) = 2\mathrm{NH_3}(g)$$

given the standard entropies

$$S_{(\mathrm{N_2}(g))}^0 = 191.6 \frac{J}{K \cdot mol}; \quad S_{(\mathrm{H_2}(g))}^0 = 130.6 \frac{J}{K \cdot mol}; \quad S_{(\mathrm{NH_3}(g))}^0 = 193.3 \frac{J}{K \cdot mol}$$

Solution:

The standard entropy change for this reaction is equal to:

$$\begin{array}{l} \Delta S^0 = \Sigma \Delta S_{(products)}^0 - \Sigma \Delta S_{(reagents)}^0 = 2 \cdot S_{(\mathrm{NH_3}(g))}^0 - \left(S_{(\mathrm{N_2}(g))}^0 + 3 \cdot S_{(\mathrm{H_2}(g))}^0\right) \\ = 2 \cdot 193.3 \frac{J}{K \cdot mol} - \left(191.6 \frac{J}{K \cdot mol} + 3 \cdot 130.6 \frac{J}{K \cdot mol}\right) \\ = -196.8 \frac{J}{K \cdot mol} \end{array}$$

Answer: the standard entropy change for this reaction is $-196.8 \frac{J}{K \cdot mol}$.

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