Solution.
ΔH(T)=ΔH(298K)+∫ΔCpdT\Delta H(T) = \Delta H(298 K) + \intop \Delta CpdTΔH(T)=ΔH(298K)+∫ΔCpdT
ΔCp=∑Cp(f)−∑Cp(i)\Delta Cp = \sum Cp(f)-\sum Cp(i)ΔCp=∑Cp(f)−∑Cp(i)
ΔCp=0.028×T−2\Delta Cp = 0.028 \times T -2ΔCp=0.028×T−2
ΔH(1000)=−42000+7T2500∣2981000−2T∣2981000=−30647.3 J=−30.6 kJmol\Delta H(1000) = -42000 + \frac{7T^2}{500}|^{1000}_{298} - 2T|^{1000}_{298} = -30647.3 \ J = -30.6 \ \frac{kJ}{mol}ΔH(1000)=−42000+5007T2∣2981000−2T∣2981000=−30647.3 J=−30.6 molkJ
Answer:
ΔH(T)=−30.6 kJmol\Delta H(T) = -30.6 \ \frac{kJ}{mol}ΔH(T)=−30.6 molkJ
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