Consider the dissociation reaction CH_4 (g) rightarrow C (s) + 2 H_2 (g) with De
ID: 966859 • Letter: C
Question
Consider the dissociation reaction CH_4 (g) rightarrow C (s) + 2 H_2 (g) with DeltaG_298degree = 50.81 kj mol^-1 and DeltaH_298degree = 74.85 kJ mol^-1. Answer the following: a. Determine the value of DeltaS_298degree for the reaction. b. What is the Delta_fH_298degree (the change in enthalpy for formation) for methane? c. Assuming that DegreeHdegree is independent of temperature, calculate K at 50degreeC. d. If DegreeC_p > 0, will DeltaHdegree increase, decrease or stay the same as the temperature is increased from 25 degreeC to 50 degreeC? Explain using the appropriate equation. e. If one mole of reaction is done in an adiabatic container starting at 298 K, will the final temperature be lower than, higher than or equal to 298 K? Explain.Explanation / Answer
1) delta G0 = delta H0 - T*delta S0
or, delta S0 = (74850 - 50810)/298 = 80.67 J/K
2) delta Hrkn = 2*delta Hf H2 + delta Hf C - delta Hf CH4
or, delta Hf CH4 = 2*0 + 0 - 74.85 = -74.85 kJ/mol
4) delta H = n*Cp*dT ; where n = moles of the gas , dT = change in temperature
Thus, increase in temperature will increase the delta H value
5) Since in adiabatic condition no heat exchange occurs .Therefore as the heat is absorbe in the reaction, the temperature will decrease.