QuestionApril 27, 2025

[13] Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical information: 4B(s)+3O_(2)(g)arrow 2B_(2)O_(3)(s) Delta H^circ =-2509.1kJ 2H_(2)(g)+O_(2)(g)arrow 2H_(2)O(l) Delta H^circ =-571.7kJ B_(2)H_(6)(g)+3O_(2)(g)arrow B_(2)O_(3)(s)+3H_(2)O(l) Delta H^circ =-2147.5kJ

[13] Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical information: 4B(s)+3O_(2)(g)arrow 2B_(2)O_(3)(s) Delta H^circ =-2509.1kJ 2H_(2)(g)+O_(2)(g)arrow 2H_(2)O(l) Delta H^circ =-571.7kJ B_(2)H_(6)(g)+3O_(2)(g)arrow B_(2)O_(3)(s)+3H_(2)O(l) Delta H^circ =-2147.5kJ
[13] Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical information: 4B(s)+3O_(2)(g)arrow 2B_(2)O_(3)(s) Delta H^circ =-2509.1kJ 2H_(2)(g)+O_(2)(g)arrow 2H_(2)O(l) Delta H^circ =-571.7kJ B_(2)H_(6)(g)+3O_(2)(g)arrow B_(2)O_(3)(s)+3H_(2)O(l) Delta H^circ =-2147.5kJ

Solution
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Answer

+35.4 \, \text{kJ} Explanation 1. Write the target reaction The standard enthalpy of formation (\Delta H_f^\circ) for B_2H_6(g) corresponds to the reaction: 2B(s) + 3H_2(g) \rightarrow B_2H_6(g) 2. Use Hess's Law Combine the given reactions to match the target reaction. Rearrange and adjust coefficients as needed. 1. Reaction 1: 4B(s) + 3O_2(g) \rightarrow 2B_2O_3(s), \Delta H^\circ = -2509.1 \, \text{kJ} - Divide by 2: 2B(s) + 1.5O_2(g) \rightarrow B_2O_3(s), \Delta H^\circ = -1254.55 \, \text{kJ} 2. Reaction 2: 2H_2(g) + O_2(g) \rightarrow 2H_2O(l), \Delta H^\circ = -571.7 \, \text{kJ} - Divide by 2: H_2(g) + 0.5O_2(g) \rightarrow H_2O(l), \Delta H^\circ = -285.85 \, \text{kJ} 3. Reaction 3: B_2H_6(g) + 3O_2(g) \rightarrow B_2O_3(s) + 3H_2O(l), \Delta H^\circ = -2147.5 \, \text{kJ} 3. Reverse Reaction 3 Reverse the third reaction to form B_2H_6(g): B_2O_3(s) + 3H_2O(l) \rightarrow B_2H_6(g) + 3O_2(g), \Delta H^\circ = +2147.5 \, \text{kJ} 4. Add reactions Add the modified reactions: 1. 2B(s) + 1.5O_2(g) \rightarrow B_2O_3(s), \Delta H^\circ = -1254.55 \, \text{kJ} 2. 3H_2(g) + 1.5O_2(g) \rightarrow 3H_2O(l), \Delta H^\circ = -857.55 \, \text{kJ} (scaled from Reaction 2) 3. B_2O_3(s) + 3H_2O(l) \rightarrow B_2H_6(g) + 3O_2(g), \Delta H^\circ = +2147.5 \, \text{kJ} Summing these: 2B(s) + 3H_2(g) \rightarrow B_2H_6(g), \Delta H^\circ = -1254.55 - 857.55 + 2147.5 = +35.4 \, \text{kJ}

Explanation

1. Write the target reaction<br /> The standard enthalpy of formation ($\Delta H_f^\circ$) for $B_2H_6(g)$ corresponds to the reaction:<br />$$ 2B(s) + 3H_2(g) \rightarrow B_2H_6(g) $$<br /><br />2. Use Hess's Law<br /> Combine the given reactions to match the target reaction. Rearrange and adjust coefficients as needed.<br /><br />1. Reaction 1: $4B(s) + 3O_2(g) \rightarrow 2B_2O_3(s)$, $\Delta H^\circ = -2509.1 \, \text{kJ}$<br /> - Divide by 2: <br /> $$ 2B(s) + 1.5O_2(g) \rightarrow B_2O_3(s), \Delta H^\circ = -1254.55 \, \text{kJ} $$<br /><br />2. Reaction 2: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$, $\Delta H^\circ = -571.7 \, \text{kJ}$<br /> - Divide by 2:<br /> $$ H_2(g) + 0.5O_2(g) \rightarrow H_2O(l), \Delta H^\circ = -285.85 \, \text{kJ} $$<br /><br />3. Reaction 3: $B_2H_6(g) + 3O_2(g) \rightarrow B_2O_3(s) + 3H_2O(l)$, $\Delta H^\circ = -2147.5 \, \text{kJ}$<br /><br />3. Reverse Reaction 3<br /> Reverse the third reaction to form $B_2H_6(g)$:<br />$$ B_2O_3(s) + 3H_2O(l) \rightarrow B_2H_6(g) + 3O_2(g), \Delta H^\circ = +2147.5 \, \text{kJ} $$<br /><br />4. Add reactions<br /> Add the modified reactions:<br />1. $2B(s) + 1.5O_2(g) \rightarrow B_2O_3(s), \Delta H^\circ = -1254.55 \, \text{kJ}$<br />2. $3H_2(g) + 1.5O_2(g) \rightarrow 3H_2O(l), \Delta H^\circ = -857.55 \, \text{kJ}$ (scaled from Reaction 2)<br />3. $B_2O_3(s) + 3H_2O(l) \rightarrow B_2H_6(g) + 3O_2(g), \Delta H^\circ = +2147.5 \, \text{kJ}$<br /><br />Summing these:<br />$$ 2B(s) + 3H_2(g) \rightarrow B_2H_6(g), \Delta H^\circ = -1254.55 - 857.55 + 2147.5 = +35.4 \, \text{kJ} $$
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