1.It is known that the D H 0 for the reaction is +624.6 kJ. SiO 2 (g) + 3C(s) It

Question

1.It is known that the DH0 for the reaction is +624.6 kJ.

SiO2(g) + 3C(s)

It is known that the DH0 for the reaction is +624.6 kJ. With excess carbon for the reaction above, the heat absorbed in the complete reaction of 3.57 grams of SiO2 will be kJ. (Report to the proper number of the significant figures) Which of the following substances is not in its standard state? Given the enthalpy changes for the following reactions, calculate for CO(g). (Report to the proper number of the significant figures) The burning of 80.3 g of SiH4 at constant pressure gives off 3790 kJ of heat. Then, the DH for the following reaction is

Explanation / Answer

ans 1)

dH0 is standard enthalpy change which is for 1 mole of reactants

moles of SiO2 = 3.57/60 = 0.0595 moles

so

change in enthalpy for 0.0595 moles = 0.0595*624.6 = 37.1637 KJ,,,,,,,,,,,,,,,,,,,,,,(4 sig fig)

ans 2)

all the above are in standadr state.

and 3)

C + O2 ----> CO2 -393.5 KJ reaction 1

CO + 1/2 O2 -----> CO2 -283.0 KJ reaction 2

C + 1/2 O2 ---> CO reaction 3

now reaction 3 = reaction 1 - reaction 2

so dH = -393.5 - (-283) = -110.5 KJ

ans 4)

mass = 80.3 gm

moles = 80.3/32 = 2.509 moles

Si H4 + 2O2 ------> Si O2 + 2H2 O

so dH for 2.509 moles = 3790 KJ

so dH = heat given by 1 mole of reactant = 3790/2.509 = 1510.5619 KJ

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