Chemistry, 13.11.2019 00:31 ridzrana02

For a particular isomer of c8h18,c8h18, the combustion reaction produces 5099.5 kj 5099.5 kj of heat per mole of c8h18(g)c8h18(g) consumed, under standard conditions. c8h18(g)+252o2(g)⟶8co2(g)+9h2o(g)δh ∘rxn=−5099.5 kj/mol c8h18(g)+252o2(g)⟶8co2(g)+9h2o(g)δh rxn°=−5099.5 kj/mol what is the standard enthalpy of formation of this isomer of c8h18(g)

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Answer from: lmimms02

Answer : The standard enthalpy of formation of this isomer of $C_8H_{18}$ is -222 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction of combustion of $C_8H_{18}$ follows:

$C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightleftharpoons 8CO_2(g)+9H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_8H_{18})}\times \Delta H^o_f_{(C_8H_{18})})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(C_8H_{18}(g))}=?\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\\\Delta H^o_{rxn}=-5099.5kJ/mol$

Putting values in above equation, we get:

$-5099.5=[(8\times -393.5)+(9\times -241.5)]-[(1\times \Delta H^o_f_{(C_8H_{18})}))+(\frac{25}{2}\times 0)]$

$\Delta H^o_f_{(C_8H_{18})})=-222kJ/mol$

Therefore, the standard enthalpy of formation of this isomer of $C_8H_{18}$ is -222 kJ/mol

Answer from: Quest

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

Answer from: Quest

You use the atomic number to find out how many protons and neutrons there are in the atom. you use the atomic mass to find out how many protons and electrons there are in the atom.

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