Chem Project

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  1.) Using standard Thermodynamic values, calculate the enthalpy of the reaction of the combustion of methane gas with oxygen gas to form carbon dioxide and liquid water. Chemical Equation: CH 4 ( g  ) + 2 O 2 ( g  ) => CO 2 ( g  ) + 2 H 2 O( l  ) + heat given : ΔH f   H 2 O( l  ) = - 285.83 kJ/mol , ΔH f   CO 2 (  g  ) = -393.51 kJ/mol, ΔH f   CH 4 (  g  ) = -74.87 kJ/mol , ΔH f   O 2 (  g  ) = 0 kJ/mol Solution: ΔH rxn  =[2 * ΔH f   H 2 O( l  ) + ΔH f   CO 2 (  g  ) ]+ [2 * ΔH f   O 2 (  g  ) + ΔH f   CH 4 (  g  )] ΔH rxn  = [2*(-285.83 kJ) + (-393.51 kJ)]  –   [(2*0 kJ) + (-74.87 kJ)] ΔH rxn  = -890.3 kJ/mol answer: Therefore the enthalpy of the reaction is -890.3 kJ/mol. 2.) If 1.6g of CH 4  reacts with oxygen gas to form water and carbon dioxide what is the change in entropy for the universe? Reaction Equation: CH 4  + 2O 2  -> CO 2  + 2H 2 O solution : ΔS   System   =ΣΔS Products  –    ΣΔS Reactants   ΔS   System =[(.21374 kJ/mol)+(2* .06995 kJ/mol)]-[(2*.20507 kJ/mol)+( .18626 kJ/mol)] ΔS   System = -.24276 kJ/mol ΔH   System   =ΣΔH Products  –    ΣΔH Reactants   ΔH   System  = [( -393.509 kJ/mol)+(2* -285.83 kJ/mol)]-[(2*0)+( -74.87 kJ/mol)] ΔH   System  = -890.229 kJ/mol ΔS   Surroundings =ΔH System  /T ΔS   Surroundings = -890.229/298 ΔS   Surroundings = -2.9873 kJ/mol Δ S Universe = ΔS   Surroundings  –ΔS   System   ΔS   Universe = -2.9873 kJ/mol  –   (-.24276 kJ/mol) ΔS   Universe  = -2.745 kJ/mol answer: Therefore the change for the entropy of the universe is -2.745 kJ/mol   3.) Find the change is Gibbs Free Energy for the reaction of hydrochloric acid and sodium hydroxide to form liquid water and sodium chloride at 31 C. First you must write the chemical equation for the reaction: HCl(aq) + NaOH(aq) ->H 2 O( l  ) +  NaCl. ΔH Rxn   =ΣΔH Products  –    ΣΔH Reactants   ΔH Rxn =[(-285.8+(-411.54)) kJ/mol]-[(-167.16 + (-470.1) kJ/mol] ΔH Rxn  = -60.05 kJ/mol ΔS Rxn   =ΣΔS Products  –    ΣΔS Reactants   ΔS Rxn =[(0.06991+ 0.07238) kJ/molK]-[0.0565 + 0.0482 kJ/molK] ΔS Rxn  = 0.03759 kJ/molK Kelvin = Celsius + 273.15 =31 + 273.15 = 304.15 K ΔG=ΔH - TΔS.   ΔG= -60.05 KJ/mol  –   304.15 K *(0.03759 KJ/molK) = -71.4830 4. ) There is a house hold heater that operates at 4 V and at 35 Ω and is used to heat up 15g of copper wire. The specific heat capacity of copper is 24.440 J/mol/K. How much time is required to increase the temperature from 25˚C to 69˚C?   It is important to know the equation in circuitry that calculates power: P=V 2 /R, which is derived from the equation V=IR. We will also be using q=mC s ΔT.   solution: P=V 2 /R   P=(4) 2 /35 P=.457 J/s q=mC s ΔT  q=(15)(24.440)(69-25) q= 16130.4 J We now know how many joules of heat must be added to the copper wire to increase the temperature and we know how many joules of energy are given off by the heater per second. We divide to find the number of seconds. Time=(16130.4 J)/(0.457 J/s) = 35296.3 seconds    5.)   What is the equilibrium constant for the formation of N 2 O 4  gas from NO 2  gas molecules? The temperature of the reaction is 310.5K.  First, the balanced equation must be written: 2NO 2(g)  -> N 2 O 4(g)  solution: K=e^[- ΔG/RT]  K is the equilibrium constant, e is the numerical value 2.718, ΔG is the change in Gibbs free energy in J/mol R is the gas constant, T is the temperature in K ΔH   System   =ΣΔH Products  –    ΣΔH Reactants   ΔH   System =[9.08 kJ/mol]-[2*33.1 kJ/mol] ΔH   System  = -57.12 kJ/mol ΔS   System   =ΣΔS Products  –    ΣΔS Reactants   ΔS   System =[.30438 kJ/molK]-[2*.24004 kJ/molK] ΔS   System  = -.1757 kJ/molK Also, the T used is not room temperature, but the temperature given in the problem  –   the temperature at which the reaction takes place. ΔG=ΔH - TΔS   ΔG=[ -57.12]-310.5[-0.1757] ΔG  = -2.565 kJ/mol or -2565 J/mol Once ΔG is calculated the srcinal equation can be used to solve for k. K=e^[- ΔG/RT]  K=e^[2565/8.314*310.5] K= 2.701 6) what is thermodynamics? answer: Thermodynamics  is a branch of  natural science concerned with heat and its relation to energy and work. It defines macroscopic variables (such as temperature, internal energy, entropy,  and pressure) that characterize materials and radiation, and explains how they are related and by what laws they change with time.  7.) What are the laws of thermodynamics? and define each. answer:    Zeroth law of thermodynamics:   If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other.  This statement implies that thermal equilibrium is an equivalence relation on the set of  thermodynamic systems under consideration. Systems are said to be in thermal equilibrium with each other if spontaneous molecular thermal energy exchanges between them do not lead to a net exchange of energy.    First law of thermodynamics:  The increase in internal energy of a closed system is equal   to the difference of the heat supplied to the system and the work done by it: ΔU = Q –   W    The first law of thermodynamics asserts the existence of a state variable for a system, the internal energy, and tells how it changes in thermodynamic processes. The law allows a given internal energy of a system to be reached by any combination of heat and work    Second law of thermodynamics:   Heat cannot spontaneously flow from a colder location   to a hotter location.   The second law of thermodynamics is an expression of the universal principle of dissipation of kinetic and potential energy observable in nature. The second law is an observation of the fact that over time, differences in temperature, pressure, and chemical potential tend to even out in a  physical system that is isolated from the outside world. Entropy is a measure of how much this    process has progressed. The entropy of an isolated system that is not in equilibrium tends to increase over time, approaching a maximum value at equilibrium.    Third law of thermodynamics:   As a system approaches absolute zero the entropy of the  system approaches a minimum value.  The third law of thermodynamics is a statistical law of nature regarding entropy and the impossibility of reaching absolute zero of temperature. This law provides an absolute reference  point for the determination of entropy. The entropy determined relative to this point is the absolute entropy. Alternate definitions are, the entropy of all systems and of all states of a system is smallest at absolute zero, or equivalently it is impossible to reach the absolute zero of temperature by any finite number of processes .

Astm e 1300

Jul 23, 2017


Jul 23, 2017
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