If a gold ring with a mass of 5.50 g changes in temperature from 25.0 to 28.0 de
ID: 473456 • Letter: I
Question
If a gold ring with a mass of 5.50 g changes in temperature from 25.0 to 28.0 degree c, how much heat has it absorbed (C = 0.129 J g^-1 K^-1) Briefly define heat capacity. C and specific heat capacity, C_s Calculate the standard enthalpy of formation of acetylene (C_2H_2) from its elements: 2 C(graphite) + H_2(g) rightarrow C_2H_2 (g) The equations for each step and the corresponding enthalpy changes are: C (graphite) + O_2(g) rightarrow CO_2 (g) delta H^o _rxn = 393.5 kJ/mol H_2(g) + 1/2 O_2(g) rightarrow H_2O(l) delta H^o _rxn = 285.8 kJ/mol 2 C_2H_2(g) + 5 O_2(g) rightarrow 4 CO_2(g) + 2 H_2O(l) delta H^o _rxn = 2598.8 kJ/mol The thermite reaction involves aluminum and iron(II) oxide 2Al(s) + Fe_2O_3(s) rightarrow Al_2O_3(s) + 2 Fe(l) This reaction is highly exothermic and the liquid iron formed is used to weld metals. Calculate the heat released in kilojoules per gram of Al reacted with Fe_2O_3. The delta H for Fe(l) is 12.40 kJ/mol. Given: deltaH^0_f for Fe_2O_3(s) = 822.2 kJ/mol delta H^0 _t for Al_2O_3(s) = -1669.8 kJ/molExplanation / Answer
Q1
a)
Apply Q equation
Q = m*C*(Tf-Ti)
substitute data
Q = 5.5*0.129*(28-25)
Q = 2.1285 J
b)
heat capacity is the amount of heat a certain object can get until the object increases by 1 °C
specific heat capacity, is the amount of heat a certain object of unit mass ( i.e. 1 kg) can get in order to icnrease 1°C its temperature
c)
HRxn = hess law
for C2H2:
from reaction (3), divide by 2
C2H2 + 5/2O2 = 2CO2 + H2O H = -2598.8/2 = -1299.4 kJ
invert (3)
2CO2 + H2O = C2H2 + 5/2O2 H = 1299.4 kJ
multiply rxn 1 by 2
2C(graphite) + 2O2 = 2CO2 H = -393.5*2 = -787 kJ
we need to cancel H2O , so leave reaction 2 as it is
H2 + 1/2 O2 = H2O H = -285.8
add all data
HTtotal = 1299.4 +-787 + -285.8 = 226.6 kJ/mo
d)
heat released when 1 g of Al reacts
mol of Al = mass/MW = 1/26.98 = 0.03706 mol of Al
basis is:
HRxn = Hprod -Hreact = (-1669.8 + 2*12.40)-(2*0+-822.2) = -822.8 kJ/rxn
then
2 mol of Al = 822.8 kJ
0.03706 mol of Al = 822.8*0.03706 = 30.492968 KJ will be released
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