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This question has multiple parts and is one question on my OWL V2 Chemistry two

ID: 520953 • Letter: T

Question

This question has multiple parts and is one question on my OWL V2 Chemistry two homework. Any help would be appreciated . I'm learning this for a test upcoming next week.

Consider the reaction

CS2(g)+4H2(g) >< is in equilibrium to form CH4(g)+2H2S(g)

Calculate delta H ,delta S , and delta G at 25°C for this reaction

Reactants/ products    delta Hf(kJ/mol) delta S

CS2(g)                         116.9                       237.9

H2(g)                            0                             130.6

CH4(g)                          -74.87                    186.1

H2S(g)                            -20.50                      205.6

b. Assume delta H and delta S are constant with respect to a change of temperature. Now calculate delta G at 450°C

c.

Compare the two values of delta G . What can you say about the spontaneity of the reaction at 25°C and at 450°C?Pick one

The reaction will be nonspontaneous at both 25°C and 450°C.

The reaction will be spontaneous at both 25°C and 450°C.

The reaction will be nonspontaneous at 25°C, and it will be spontaneous at 450°C.

The reaction will be spontaneous at 25°C, and it will be nonspontaneous at 450°C.

Explanation / Answer

Calculate delta H ,delta S , and delta G at 25°C for this reaction

a)

dH = Hproducts - Hreactants

dH = (CH4 + 2*H2S) - (CS2 + 4H2)

dH = -74.87 + 2*-20.5 - (116.9 + 4*0) = -232.77 kJ/mol

apply:

dS = Sproducts - Sreactants

dS = (CH4 + 2*H2S) - (CS2 + 4H2)

dS = 186.1 + 2*205.6 - (237.9 + 4*130.6) = -163 J/molK

finally

dG = dH - T*dS = -232.77- 298 * (-163 /1000) = -184.196 kJ/mol

b)

T = 450°C = 723 K

dG = dH - T*dS

dG = dH - T*dS = -232.77- 723 * (-163 /1000) = -114.921 kJ/mol

c)

both are negative, so they are both spontaneous at both T

The reaction will be spontaneous at both 25°C and 450°C.

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