The reaction of sodium metal with liquid water is very exothermic and proceeds a

Question

The reaction of sodium metal with liquid water is very exothermic and proceeds according to the following chemical equation. 2 Na(s) + 2 H_2O(l) rightarrow 2 NaOH(aq) + H_2(g) DeltaHdegree_rxn = -183.8 kJmiddotmol^-1 Determine the work performed when a 0.434g piece of sodium is reacted with excess water at 18 degree C and 1.00 atm. [1 Lmiddotatm = 101.325J] A) 9.44 J B) 9.44 J C) -22.8 J D) 45.6 J Given the following date: C_3H_6(g) + H_2(g) rightarrow C_3H(g) DeltaHdegree_rxn = -124 kJmiddotmol-1 C_3H_8(g) + 5 O_2(g) rightarrow 3 CO_2(g) + 4 H_2O(l) DeltaHdegree_rxn = -2220 kJmiddotmol-1 H_2(g) + 1/2 O_2(g) rightarrow H_2O(l) DeltaHdegree_rxn = -286 kJmiddolmol-1 Determine the value of DeltaHdegree_rxn for C_3H_6(g) + 9/2 O_2(g) rightarrow 3 CO_2(g) + 3 H_2O(l) A) -1772 kJ/mol B) -2058 kJ/mol C) -3488 kJ/mol D) -2630 kJ/mol Charcoal is composed of mostly carbon. Assuming that a lump of charcoal is entirely composed of carbon, and given the following chemical equation, C(s) + O_2(g) rightarrow CO_2(g) DeltaHdegree_rxn = -393.5 kJmiddotmol-1 Determine the energy released as heat when 5.2 kg of charcoal are burned in the presence of excess oxygen at constant pressure. A) 1.70 middot 10^-5 kJ B) -1.70 middot 10^-5kJ C) 24.6 kJ D) -1.10 kJ Determine DeltaHdegree_rxn for the reaction described by the chemical equation CaCO_3(s) rightarrow CaO(s) + CO_2(g) given the following DeltaHdegreet values (in kJmiddotmol-1):CaCO_3(s), -1206.9; CaO(s), -635.6; A) 813.4 kJmiddotmol^-1 B) 177.8 kJmiddotmol^-1 C) 2236.0 kJmiddotmol^-1D) -964.8 kJmiddotmol^-1 Use average molar bond energies to estimate DeltaHmiddot_rxn for the equation 3 H_2(g) + N_2(g) rightarrow 2 NH_3(g) Given the following H_bond values (in kJ/mol): H-H 435; N N 945; N-H 390 A) 83 kJmiddotmol^-1 B) -990 kJmiddotmol^-1 C) -90 kJmiddotmol^-1 D) 1860 kJmiddotmol^-1 The specific heat capacity of silver is 0.24 Jmiddotg^-1middotdegreeC^-1. What amount of energy is required to raise the temperature of 12.4 grams of silver from 20 degreeC to 35 degreeC at constant pressure? A) 0.24 J B) 17 J C) 45 J D) 3.4 J

Explanation / Answer

25)

we know that

moles = mass / molar mass

so

moles of Na = 0.434 / 23 = 0.01887

now

2 Na + 2H20 --> 2 NaOH + H2

moles of H2 formed = 0.5 x moles of Na reacted

moles of H2 formed = 0.5 x 0.01887

moles of H2 formed = 9.435 x 10-3

now

PV = nRT

1 x V = 9.435 x 10-3 x 0.0821 x 291

V = 0.2254 L

now

work done = P x dV

work done = 1 x 0.2254

work done = 0.2254 L atm

= 0.2254 x 101.315

= 22.8

so

work performed is -22.8 J

the answer is C) -22.8

26)

a) C3H6 + H2 ---> C3H8

b) C3H8 + 5 02 ---> 3 C02 + 4H20

c) H2 + 0.5 02 ---> H20

d) C3H6 + 4.5 02 --> 3 C02 + 3 H20

we can see that

d = a + b - c

so

dHd = dHa + dHb - dHc

dHd = -124 -2220 + 286

dHd = -2058 kJ/mol

so

the answer is B) -2058 kJ/mol

27)

we know that

moles = mass / molar mass

so

moles of C = 5.2 x 1000 / 12

moles of C = 433.333

now

1 mole of C ---> -393.5

433.33 mole of C ---> y

y = 433.33 x -393.5

y = -1.7 x 10^5 kJ

so

the answer is B) -1.7 x 10^5 kJ

28)

we know that

dHrxn = dHof products - dHof reactants

so

dHrxn = dHof CaO + dHof C02 - dHof CaC03

dHrxn = -635.6 - 393.5 + 1206.9

dHrxn = 177.8 kJ/mol

so

the answer is B) 177.8 kJ/mol

29)

we know that

dHrxn = bond energies of reactants - bond energies of products

dHrxn = ( 3 x BE of H-H) + ( BE of N=N) - ( 6 x BE of N-H)

dHrxn = ( 3 x 435 ) + ( 945) - ( 6 x 390)

dHrxn = -90 KJ/mol

so

the answer is C) -90 kJ/mol

30)

we know that

heat = mass x specific heat x temp change

Q = m x s x dT

Q = 12.4 x 0.24 x ( 35-20)

Q = 44.64

so

45 J of energy is required

the answer is C) 45 J

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