The heat of reaction for N2+3H2⇔2NH3 at 27°C is -91.94 kJ. Its value at 50°C is _____. The molar heat capacities at constant P and 27°C for N2, H2 and NH3 are 28.45, 28.32 and 37.07 joule mol-1 respectively.
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answer is -92.84.
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Detailed Solution
N2+3H2→2NH3ΔH=−91.94kJ at 300K Also, ΔH2−ΔH1=ΔCpT2−T1ΔCpJmor−1=2×CPNH3−CPN2−3×CPH2=2×37.07−28.45−3×28.32ΔCp=−39.27J∴ΔH2+91940=−39.27×(323−300)ΔH2=−92843.2J=−92.843kJ