Heat of Reaction versus Temperature (Kirchhoff’s Law): 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2016, Q3

Question 3: Heat of Reaction versus Temperature (Kirchhoff’s Law) (Part A — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\Delta H^\circ_R(298\ \text{K}) = -21.59$ kcal/mol. Heat-capacity coefficients (the printed units read “kcal/mol·K”, but the magnitude of $a\approx29$ shows these are in J·mol⁻¹·K⁻¹):

Component	$a$	$b\times10^{3}$	$c\times10^{6}$	$d\times10^{9}$
CO	29.03	−2.82	11.64	−4.71
H₂	28.61	1.02	−0.15	0.77
CH₃OH	21.14	70.84	25.87	−28.50

Find. $\Delta H_R(T)$ as an explicit polynomial in $T$.

Approach. Form $\Delta C_p = C_{p,\text{CH}_3\text{OH}}-C_{p,\text{CO}}-2\,C_{p,\text{H}_2}$, integrate Kirchhoff’s equation $\text{d}(\Delta H)/\text{d}T=\Delta C_p$, and fix the integration constant $\Delta H_0$ from the 298 K datum (after converting kcal to J).

Coefficient differences. With $\Delta\theta = \theta_{\text{CH}_3\text{OH}}-\theta_{\text{CO}}-2\theta_{\text{H}_2}$ for each of $a,b,c,d$: $$\Delta a=-65.11,\quad \Delta b=71.62\times10^{-3},\quad \Delta c=14.53\times10^{-6},\quad \Delta d=-25.33\times10^{-9}.$$
Convert the datum to joules. This is the step that most often goes wrong: the $C_p$’s are in J, so $$\Delta H^\circ_R(298.15)=-21.59\ \text{kcal/mol}\times4184\ \tfrac{\text{J}}{\text{kcal}}=-90{,}333\ \text{J/mol}.$$
Integrate Kirchhoff’s law. $\displaystyle \Delta H_R(T)=\Delta H_0+\Delta a\,T+\tfrac{\Delta b}{2}T^2+\tfrac{\Delta c}{3}T^3+\tfrac{\Delta d}{4}T^4.$ Evaluating the polynomial part at 298.15 K gives $-16{,}151$ J, so $$\Delta H_0=-90{,}333-(-16{,}151)=\boxed{-74{,}182\ \text{J/mol}}.$$
Empirical equation. Substituting the constants (with $T$ in K, $\Delta H_R$ in J/mol): $$\boxed{\;\Delta H_R(T)=-74{,}182-65.11\,T+0.03581\,T^2+4.843\times10^{-6}\,T^3-6.333\times10^{-9}\,T^4\;}$$
Check. At 298.15 K it returns $-90{,}333$ J $=-21.59$ kcal ✓. As a sample, at 500 K it gives $-97{,}574$ J/mol $=-23.32$ kcal/mol (the reaction becomes slightly more exothermic as $T$ rises, since $\Delta C_p<0$).

Quantity	Result
$\Delta H_0$ (integration constant)	−74,182 J/mol
$\Delta H_R(T)$	−74182 − 65.11$T$ + 0.03581$T^2$ + 4.843e−6$T^3$ − 6.333e−9$T^4$ (J/mol)
Check: $\Delta H_R(500\,\text{K})$	−23.32 kcal/mol

Concept & Theory

Topic. Temperature dependence of the heat of reaction — Kirchhoff’s law.

Key relations. $\left(\dfrac{\partial \Delta H_R}{\partial T}\right)_P=\Delta C_p$; $\Delta C_p=\sum_i\nu_i C_{p,i}$; integrate to $\Delta H_R(T)=\Delta H_0+\int\Delta C_p\,\text{d}T$.

Why this works. Enthalpy is a state function, so heating reactants and products along their $C_p$ curves and reacting at any $T$ must give the same $\Delta H$ as reacting at 298 K and heating the products — the difference is exactly $\int\Delta C_p\,\text{d}T$. The one integration constant is pinned by the single measured standard value.

Common pitfall. Mixing energy units — the $C_p$ coefficients here are joules while the datum is kilocalories; failing to multiply the datum by 4184 puts $\Delta H_0$ off by exactly that factor. Also watch the $10^{3},10^{6},10^{9}$ column multipliers.

Source. Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering §23 (heats of reaction, effect of temperature); Smith, Van Ness & Abbott §4.6.

Question 3 of 6: Heat of Reaction versus Temperature (Kirchhoff’s Law)

Question 3: Heat of Reaction versus Temperature (Kirchhoff’s Law) (Part A — equal value)