Standard enthalpy of formation of NH3 (g) is -46.11 kJ mol-1 at 298 K. Given the heat capacity data below, calculate the standard enthalpy of formation at 1200 K. CP,m (H2 (g))/ J mol-1 K-1 = 29.1 – (0.84 x 10-3 K-1)T CP,m (N2 (g))/ J mol-1 K-1 = 26.98 + (5.9 x 10-3 K-1)T CP,m (NH3 (g))/ J mol-1 K-1 = 25.9 + (33 x 10-3 K-1)T (-53.1 kJ/mol)
The Correct Answer and Explanation is:
To calculate the standard enthalpy of formation of ammonia (NH₃(g)) at 1200 K, we use the temperature correction method based on heat capacity data. The relationship used is:ΔHf∘(1200 K)=ΔHf∘(298 K)+ΔH∘(1200 K)−ΔH∘(298 K)\Delta H_f^\circ(1200\ \text{K}) = \Delta H_f^\circ(298\ \text{K}) + \Delta H^\circ(1200\ \text{K}) – \Delta H^\circ(298\ \text{K})ΔHf∘(1200 K)=ΔHf∘(298 K)+ΔH∘(1200 K)−ΔH∘(298 K)
Where the change in enthalpy is calculated from:ΔH=∫2981200∑νCP,m∘(products) dT−∫2981200∑νCP,m∘(reactants) dT\Delta H = \int_{298}^{1200} \sum \nu C_{P,m}^\circ(\text{products})\ dT – \int_{298}^{1200} \sum \nu C_{P,m}^\circ(\text{reactants})\ dTΔH=∫2981200∑νCP,m∘(products) dT−∫2981200∑νCP,m∘(reactants) dT
Step 1: Write the formation reaction
12N2(g)+32H2(g)→NH3(g)\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)21N2(g)+23H2(g)→NH3(g)
Step 2: Heat capacity equations (in J/mol·K)
- CP,m(H2)=29.1−0.00084TC_{P,m}(H_2) = 29.1 – 0.00084TCP,m(H2)=29.1−0.00084T
- CP,m(N2)=26.98+0.0059TC_{P,m}(N_2) = 26.98 + 0.0059TCP,m(N2)=26.98+0.0059T
- CP,m(NH3)=25.9+0.033TC_{P,m}(NH_3) = 25.9 + 0.033TCP,m(NH3)=25.9+0.033T
Step 3: Compute enthalpy change from 298 K to 1200 K
Integrate CP(T)C_P(T)CP(T) for each species:
For NH₃:ΔHNH3=∫2981200(25.9+0.033T) dT=25.9(T)+0.033T22⇒[25.9T+0.0165T2]2981200\Delta H_{NH_3} = \int_{298}^{1200}(25.9 + 0.033T)\,dT = 25.9(T) + 0.033\frac{T^2}{2} \Rightarrow [25.9T + 0.0165T^2]_{298}^{1200}ΔHNH3=∫2981200(25.9+0.033T)dT=25.9(T)+0.0332T2⇒[25.9T+0.0165T2]2981200=(25.9×1200+0.0165×12002)−(25.9×298+0.0165×2982)=48828+23760−7718.2−1465.4=63304.4J/mol= (25.9×1200 + 0.0165×1200^2) – (25.9×298 + 0.0165×298^2) = 48828 + 23760 – 7718.2 – 1465.4 = 63304.4 J/mol=(25.9×1200+0.0165×12002)−(25.9×298+0.0165×2982)=48828+23760−7718.2−1465.4=63304.4J/mol
For 12N2\frac{1}{2}N_221N2:ΔH=0.5×[26.98T+0.00295T2]2981200=0.5×[(26.98×1200+0.00295×12002)−(26.98×298+0.00295×2982)]=0.5×(32376+4248−8036.04−261.2)=0.5×(36624−8297.24)=14163.4J/mol\Delta H = 0.5 \times \left[26.98T + 0.00295T^2\right]_{298}^{1200} = 0.5 \times [(26.98×1200 + 0.00295×1200^2) – (26.98×298 + 0.00295×298^2)] = 0.5 \times (32376 + 4248 – 8036.04 – 261.2) = 0.5 \times (36624 – 8297.24) = 14163.4 J/molΔH=0.5×[26.98T+0.00295T2]2981200=0.5×[(26.98×1200+0.00295×12002)−(26.98×298+0.00295×2982)]=0.5×(32376+4248−8036.04−261.2)=0.5×(36624−8297.24)=14163.4J/mol
For 32H2\frac{3}{2}H_223H2:ΔH=1.5×[29.1T−0.00042T2]2981200=1.5×[(29.1×1200−0.00042×12002)−(29.1×298−0.00042×2982)]=1.5×(34920−604.8−8671.8+37.3)=1.5×25680.7=38521.0J/mol\Delta H = 1.5 \times [29.1T – 0.00042T^2]_{298}^{1200} = 1.5 × [(29.1×1200 – 0.00042×1200^2) – (29.1×298 – 0.00042×298^2)] = 1.5 × (34920 – 604.8 – 8671.8 + 37.3) = 1.5 × 25680.7 = 38521.0 J/molΔH=1.5×[29.1T−0.00042T2]2981200=1.5×[(29.1×1200−0.00042×12002)−(29.1×298−0.00042×2982)]=1.5×(34920−604.8−8671.8+37.3)=1.5×25680.7=38521.0J/mol
Step 4: Total ΔH correction
ΔH=HNH3−HN2+H2=63304.4−(14163.4+38521.0)=10620J/mol=10.62kJ/mol\Delta H = H_{NH_3} – H_{N_2+H_2} = 63304.4 – (14163.4 + 38521.0) = 10620 J/mol = 10.62 kJ/molΔH=HNH3−HN2+H2=63304.4−(14163.4+38521.0)=10620J/mol=10.62kJ/mol
Step 5: Apply correction
ΔHf∘(1200 K)=−46.11 kJ/mol+10.62 kJ/mol=−35.49 kJ/mol\Delta H_f^\circ(1200\ K) = -46.11\ \text{kJ/mol} + 10.62\ \text{kJ/mol} = \boxed{-35.49\ \text{kJ/mol}}ΔHf∘(1200 K)=−46.11 kJ/mol+10.62 kJ/mol=−35.49 kJ/mol
Explanation
The standard enthalpy of formation, ΔHf∘\Delta H_f^\circΔHf∘, of a compound is the enthalpy change when one mole of the substance forms from its elements in their standard states. For NH₃(g), this value is -46.11 kJ/mol at 298 K. However, enthalpy is temperature-dependent because molecular heat capacities vary with temperature.
To determine the enthalpy of formation at 1200 K, we account for the temperature-dependent change in enthalpy using heat capacity expressions. The reaction considered is:12N2(g)+32H2(g)→NH3(g)\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)21N2(g)+23H2(g)→NH3(g)
For each species, we integrate their molar heat capacity (CP,mC_{P,m}CP,m) from 298 K to 1200 K to obtain the change in enthalpy with temperature. The integral of a linear function CP=a+bTC_P = a + bTCP=a+bT over temperature gives:ΔH=a(T2−T1)+b2(T22−T12)\Delta H = a(T_2 – T_1) + \frac{b}{2}(T_2^2 – T_1^2)ΔH=a(T2−T1)+2b(T22−T12)
By applying this to each component in the reaction and calculating the net enthalpy difference between products and reactants, we obtain the total enthalpy correction. This correction (10.62 kJ/mol) is added to the standard enthalpy of formation at 298 K to yield the value at 1200 K.
Thus, the standard enthalpy of formation of NH₃(g) at 1200 K is -35.49 kJ/mol, reflecting the increased thermal energy at elevated temperatures and reduced thermodynamic driving force for ammonia formation.
