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ΔH=-q

ΔH R1=-1.875kJ

ΔH R2=-4.678kJ

ΔH R3=-2.865kJ

 

mol R1=2.00/(22.99+16.00+1.008)=0.05 mol

mol R2=2.00/(22.99+16.00+1.008)=0.05 mol

mol R3=1.0*50.1000=0.05 mol

 

ΔH/mol R1=-1.875/0.05=-37.5 kJ/mol

ΔH/mol R2=-4.678/0.05=-93.6 kJ/mol

ΔH/mol R3=-2.865/0.05=-57.3 kJ/mol

 

-ΔH1  ΔH2  ΔH3

-37.5+-57.3=-94.8 kJ/mol

37.5+-93.6=-56.1 kJ/mol

|-94.8+93.6| / |-93.6| *100=1.28% error

Conclusion: