CaC(s) + H2O(l) > Ca(OH)2(s): triangle H = -65.2kJ

1) 155.65 kJ; 2) -59.0 kJ/mol; 3) a)

Step-by-step explanation:

Answer on Question # 55533 - Chemistry - General chemistry

Question:

1. Given the data. N2(g) + O2(g) = 2 NO(g) ΔH = +180.7 kJ; 2 NO(g) + O2(g) = 2 NO2(g) ΔH = −113.1

kJ; 2 N2O(g) = 2 N2(g) + O2(g) ΔH = − 163.2 kJ; Use Hess’s Law to calculate ΔH for the reaction

N2O(g) + NO2(g) = 3 NO(g); Show your work.

2. Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2. From the

following enthalpy of reaction data and data in Appendix C in textbook, calculate ΔHf° for CaC2(s):

CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2 (g) ΔH = −127.2kJ

3. Using average bond enthalpies, predict which of the following reactions will be most

exothermic: a) C(g) + 2 F2(g) CF4(g) b) CO(g) +3 F2(g) CF4(g) + OF2(g) c) CO2(g) + 4 F2(g) CF4(g) +

2 OF2(g)

Solution

1)

N2(g) + O2(g) = 2 NO(g) ΔH1 = +180.7 kJ x(+2)

2 NO(g) + O2(g) = 2 NO2(g) ΔH2 = −113.1 kJ x(-1)

2 N2O(g) = 2 N2(g) + O2(g) ΔH3 = − 163.2 kJ x(+1)

2N2O(g) + 2NO2(g) = 6 NO(g) ΔH = (ΔH3 + 2ΔH1 – ΔH2)

N2O(g) + NO2(g) = 3 NO(g) ΔH = (ΔH3 + 2ΔH1 – ΔH2)/2 = 155.65 kJ

2)

CaC2(s) + 2 H2O(l) = Ca(OH)2(s) + C2H2 (g) ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) - ΔHf°(CaC2);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

3)

a) C(g) + 2 F2(g) = CF4(g)

b) CO(g) +3 F2(g) = CF4(g) + OF2(g)

c) CO2(g) + 4 F2(g) = CF4(g) + 2 OF2(g)

C-F bond enthalpy 440 kJ/mol

C=O bond enthalpy in carbon dioxide 805 kJ/mol

C=O bond enthalpy in carbon monoxide 1077 kJ/mol

O-F bond enthalpy 184 kJ/mol

F-F bond enthalpy 153 kJ/mol

a) ΔHrxn = 2x153 - 4x440 = -1454 kJ – the most exothermic

b) ΔHrxn = 1077 + 3x152 - 2x184 - 4x440 = -595 kJ

c) ΔHrxn =805x2 + 4x153 - 4x440 – 2x2x184 = -274 kJ

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab  in the y coordinate 

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

  x coordinate 

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

  y coordinate 

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

  x  coordinate 

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

  y  coordinate 

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

  x  coordinate 

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

  y  coordinate 

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

  x  coordinate 

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

  y  coordinate 

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

  x  coordinate 

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

  y  coordinate 

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

  x  coordinate 

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

  y  coordinate 

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

  x  coordinate 

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

  y  coordinate 

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy