Question

In: Chemistry

Given the following thermochemical equations, 2Cu(s) + S(s) right-arrow Cu2S(s) ΔH°= –79.5 kJ S(s) + O2(g)...

Given the following thermochemical equations, 2Cu(s) + S(s) right-arrow Cu2S(s) ΔH°= –79.5 kJ S(s) + O2(g) right-arrow SO2(g) ΔH°= –297 kJ Cu2S(s) + 2O2(g) right-arrow 2CuO(s) + SO2(g) ΔH° = –527.5 kJ calculate the standard enthalpy of formation (in kJ mole–1) of CuO(s).

Multiplied by 1/2

Finally, using the thermodynamic values given and knowing how to add all the equations together to get the desire equation, calculate the value of ΔH_f° for the final formation reaction.

Expert Solution

2Cu(s) + S(s) ------------------> Cu2S(s) ΔH°= –79.5 kJ -------------------> 1

S(s) + O2(g) -------------------> SO2(g) ΔH°= –297 kJ ---------------------> 2

Cu2S(s) + 2O2(g) ------------->2CuO(s) + SO2(g) ΔH° = –527.5 kJ --------------> 3

adding 1 and 3 equations

2Cu(s) + S(s) ------------------> Cu2S(s) ΔH°= –79.5 kJ

Cu2S(s) + 2O2(g) ------------->2CuO(s) + SO2(g) ΔH° = –527.5 kJ adding two equations

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2Cu(s) + S(s) + 2O2(g) -------------> 2CuO(s) + SO2(g) ΔH° = -607Kj ---------------> 4

substract 4 and 2 equation

2Cu(s) + S(s) + 2O2(g) -------------> 2CuO(s) + SO2(g) ΔH° = -607Kj ---------------> 4

S(s) + O2(g) -------------------> SO2(g) ΔH°= –297 kJ ---------------------> 2

(-) (-) (-) +

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2Cu(s) + O2(g) -----------------> 2CuO ΔH° = -310Kj

It is divided to 2 to get formation of CuO

Cu(s) + 1/2O2(g) -----------------> CuO ΔHf° = -155Kj

queen_honey_blossom answered 2 years ago

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