Consider the Haber process: N_2(g) + 2H_2(g) rightarrow 2NH_3(g) Delta H = -46 k

Question

Consider the Haber process: N_2(g) + 2H_2(g) rightarrow 2NH_3(g) Delta H = -46 kJ/mol of NH_3 formed. If 6 g of H_2 are reacted from this reaction, how much heat is associated with the reaction? -92 kJ +92 kJ -76 kJ +76 kJ consider the fusion reaction of: NH_3(aq) NH_3(g) What is the normal boiling point of NH_3, 142 degree C 100 degree C 415 degree C 0.4 K When patm is 0.823 atm, what is the boiling point? 58.5 degree 134 degree C 407 degree C When this reaction gets equilibrium Delta G degree = Delta G Delta G = 0 K = 0 Cannot determine How much heat must be added to 42.3g of liquid water at 75 degree C to convert 105 degree C of steam water? 96 kJ 10.7 kJ 9.7 kJ 100 kJ Calculate Delta Hrxn for the following reaction based on the given data: S(s) + O_2 (g) rightarrow SO_2(g) 2SO_2(g) + O_2 (g) rightarrow 2SO_3 (g) Delta H = -198.2 -296.1 KJ -592.2 KJ 592.2 KJ

Explanation / Answer

12. moles of H2 = 6/2 = 3 mols

3/2 moles gives -46 kJ/mol heat

so 3 moles would give = -46 x 3 x 2/3 = -92 kJ of heat

So answer is,

a. -92 kJ

For the given fusion reaction,

13. Normal boiling point of NH3 is,

a. 142 oC

14. When patm = 0.823 atm, the boiling point is,

a. 58.5 oC

16. Heat required to convert 42.3 g of water at 75 oC to steam water at 105 oC would be,

q = 42.3 x 4.184 x 25 + 42.3 x 2260 + 42.3 x 2.03 x 5 = 100 kJ

So answer is,

d. 100 kJ

17. Heat of reaction,

divide first equation by 2 and invert the reaction, then add to the second equation,

dHrxn = 198.2/2 - 395.2 = -296.1 kJ

so the asnwer is,

b. -296.1 kJ

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