6. The normal boiling point of ammonia is-33C. For the process at-40C, the signs
ID: 1043070 • Letter: 6
Question
6. The normal boiling point of ammonia is-33C. For the process at-40C, the signs of ??, as, and ?G would be, respectively, b) 7. Calculate AG for the reaction C(s)CO2(g) 2CO(g) a) b) e) -120 kJ 120 k c) -257.2 k d) +257.2 k cannot determine because has not been given for C(s) 8. Given N2(g) + O2(g) N2(g) + 2O2(g) 2NO(g) ?«Go-86.60k/mol 2NO2(g) ?/G"x51 kJ/mol AG for the following reaction? NO(g) +1/202 NOB) What is 9. Which of the following statements is true about the reaction 2Be(s) + 02lg) 2BeO(s)? (A(G-580.1 kJ/mol for BeO(s) at 298 K.) a) Under standard-state conditions, the reaction is spontaneous because AGo is negative. b) Under standard-state conditions, the reaction is spontaneous because AG° is positive c) Under standard-state conditions, the reaction is nonspontaneous because AG° is negative d) Under standard-state conditions, the reaction is nonspontaneous because ?G° is positive. e) None of the above statements are true.Explanation / Answer
6) Ammonia can be liquefied by applying some pressure i.e compression process
at a temperature below -33 o C ammonia runs in to liquid at atmospheric pressure, the given temperature of -40o C is less than -33o C so
delta H will be +ve
delta S will be -ve as the randomness decreases as the gas turns in to liquid.
dealt G = delta H - T delta S = -ve as this is a spontaneous process at the given temperature.
Option e is the right answer.
7) delta Grxn = Sum of the deltaG of products - sum of delta G of reactants
= [2 x (-137.2 kJ/mol)]- [-394.4 kJ/mol+ 0 kJ/mol] = + 120 kJ
option b is the right answer
** answering only first two questions due to policy and time constraints , please post the questions separately. Sorry to dissapoint you.
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