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18 Figure 7 shows the displayed formulae equation for the reaction of nitrogen with hydrogen. Figure 7 [ mathrm(N) equiv mathrm(N)+mathrm(SH)-mathrm(H) longrightarrow 2 mathrm(H)-mathrm(N)-mathrm(H) ] H In the reaction the energy released forming new bonds is 93 mathrm(kJimol) greater than the energy needed to break existing bonds. Table 3 shows bond energies. Table 3 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 0 longdiv ( 5 ) cdot 3 Calculate the bond energy mathrm(X) for the mathrm(H)-mathrm(H) bond. [5 marks] Use Figure 7 and Table 3. omega arrow

Question

18
Figure 7 shows the displayed formulae equation for the reaction of nitrogen with hydrogen.
Figure 7
[
mathrm(N) equiv mathrm(N)+mathrm(SH)-mathrm(H) longrightarrow 2 mathrm(H)-mathrm(N)-mathrm(H)
]
H
In the reaction the energy released forming new bonds is 93 mathrm(kJimol) greater than the energy needed to break existing bonds.
Table 3 shows bond energies.
Table 3

 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) 
 Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 



 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) 
 Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 


 0 longdiv ( 5 ) cdot 3 Calculate the bond energy mathrm(X) for the mathrm(H)-mathrm(H) bond.
[5 marks]
Use Figure 7 and Table 3.
 omega arrow

18 Figure 7 shows the displayed formulae equation for the reaction of nitrogen with hydrogen. Figure 7 [ mathrm(N) equiv mathrm(N)+mathrm(SH)-mathrm(H) longrightarrow 2 mathrm(H)-mathrm(N)-mathrm(H) ] H In the reaction the energy released forming new bonds is 93 mathrm(kJimol) greater than the energy needed to break existing bonds. Table 3 shows bond energies. Table 3 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 Bond & mathrm(N)=mathrm(N) & mathrm(H)-mathrm(H) & mathrm(N)-mathrm(H) Bond energy in mathrm(kJ) / mathrm(mol) & 945 & mathrm(X) & 391 0 longdiv ( 5 ) cdot 3 Calculate the bond energy mathrm(X) for the mathrm(H)-mathrm(H) bond. [5 marks] Use Figure 7 and Table 3. omega arrow

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RosalindElite · Tutor for 8 years

Answer

The bond energy \(X\) for the \(H-H\) bond is \(368.5 kJ/mol\).

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## Step 1: <br />Firstly, we need to understand the reaction given. The reaction is:<br />\[<br />\mathrm{N}+3 \mathrm{~N}-\mathrm{H} \longrightarrow 2 \mathrm{~N}<br />\]<br />This means that one molecule of nitrogen (N) reacts with three molecules of hydrogen (H) to form two molecules of ammonia (NH3).<br /><br />## Step 2: <br />Next, we need to calculate the total energy needed to break the bonds in the reactants. According to Table 3, the bond energy of N=N is 945 kJ/mol and the bond energy of N-H is 391 kJ/mol. Since there is one N=N bond and three N-H bonds in the reactants, the total energy needed to break the bonds in the reactants is:<br />### \(945 kJ/mol + 3 \times 391 kJ/mol = 2118 kJ/mol\)<br /><br />## Step 3: <br />We are given that the energy released forming new bonds is 93 kJ/mol greater than the energy needed to break existing bonds. This means that the total energy released when the new bonds are formed is:<br />### \(2118 kJ/mol + 93 kJ/mol = 2211 kJ/mol\)<br /><br />## Step 4: <br />Finally, we need to calculate the bond energy of the H-H bond. According to the reaction, there are six H-H bonds in the products (two in each NH3 molecule). Therefore, the bond energy of the H-H bond is:<br />### \(X = \frac{2211 kJ/mol}{6} = 368.5 kJ/mol\)
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