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Difference between Internal Energy and Enthalpy

Becker and Ru...

Question

Becker and Ruth measured the heat evolved in the following processes at $20^{o} C$ : (1) $1$ mole of solid ${({C O O N H}_{4})}_{2} H_{2} O$ is burned in oxygen, (2) $1$ mole of solid ${(C O O H)}_{2} {(H_{2} O)}_{2}$ is burned in oxygen, (3) $1$ mole of solid ${({C O O N H}_{4})}_{2} H_{2} O$ is dissolved in a large excess of water, (4) $1$ mole of solid ${(C O O H)}_{2} {(H_{2} O)}_{2}$ is dissolved in a large excess of water, (5) $1$ mole of oxalic acid in dilute solution is neutralized with gaseous ammonia. They found (1) $189.86 k c a l$ , (2) $53.10 k c a l$ , (3) $- 11.47 k c a l$ , (4) $- 8.62 k c a l$ , (5) $43.13 k c a l$ ; (1) and (2) were measured at constant volume, the others at constant pressure. The end products of (1) and (2) were nitrogen, carbon-di-oxide and water. The heat of formation for $1$ mole of water from the elements had previously been determined as $68.35 k c a l$ at constant pressure and $20^{o} C$ . Find the change in enthalpy when $1$ mole of ${N H}_{3}$ is formed from the elements at $20^{o} C$ (only magnitude in nearest integer in kcal)

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Solution

$R_{1} : (C O O N H_{4})_{2} H_{2} O (s) + 2 O_{2} (g) \to N_{2} (g) + 2 C O_{2} (g) + 5 H_{2} O (l); Δ Q_{1} = 189.86 k c a l$
$R_{2} : (C O O H)_{2} (H_{2} O)_{2} (s) + \frac{1}{2} O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l); Δ Q_{2} = 53.10 k c a l$
$R_{3} : (C O O N H_{4})_{2} H_{2} O (s) \to (C O O N H_{4})_{2} (a q) + H_{2} O; Δ Q_{3} = - 11.47 k c a l$
$R_{4} : (C O O H)_{2} (H_{2} O)_{2} (s) \to (C O O H)_{2} (a q) + 2 H_{2} O (l); Δ Q_{4} = - 8.62 k c a l$
$R_{5} : (C O O H)_{2} (a q) + 2 N H_{3} (g) \to (C O O N H_{4})_{2} (a q); Δ Q_{5} = 43.13 k c a l$
$R_{6} : H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ Q_{6} = 68.35 k c a l$
Formation of ammonia from the elements
$R_{7} : \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g); Δ Q_{7}$
rearranging there reactions give
$2 R_{7} = - R_{5} - R_{4} + R_{3} + R_{2} - R_{1} + 3 R_{5}$
Therefore, according to Hess's Law of constant heat summation
$2 Δ Q_{7} = - Δ Q_{5} - Δ Q_{4} + Δ Q_{3} + Δ Q_{2} - Δ Q_{1} + 3 Δ Q_{6}$
$2 Δ Q_{7} = - 43.13 + 8.62 - 11.47 + 53.10 - 189.86 + 3 \times 68.35 = 22.31 k c a l$
$\Rightarrow Δ Q_{7} = \frac{22.31}{2} = 11.15 k c a l$
Answer = 11

Note: $R_{5}$ reaction happens in two steps
$N H_{3} (g) + H_{2} O (l) \to N H_{4} O H (a q)$
$(C O O H)_{2} (a q) + N H_{4} O H (a q) \to (C O O N H_{4})_{2} (a q) + H_{2} O (l)$

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