Answer:
∆H0 = -222kJ/mol
Explanation:
Using Hess's law, we can find the ΔH of a reaction from the sum of another related reactions as follows:
Using the reactions:
(1) C(s) + O2(g) → CO2(g) ∆H0= -394 KJ/mol
(2) CO(s) + 1/2 O2(g) → CO2(g) ∆H0= -283 KJ/mol
Twice (1):
2C(s) + 2O2(g) → 2CO2(g) ∆H0= 2*-394 KJ/mol = -788kJ/mol
The inverse reaction of (2):
-(2) CO2(g) → CO(g) + 1/2 O2(g) ∆H0= 283 KJ/mol
Twice this reaction:
2*-(2) 2CO2(g) → 2CO(s) + O2(g) ∆H0= 2*283 KJ/mol= 566kJ/mol
Now, the sum of 2*(1) - 2*(2) produce:
2C(s) + 2O2(g) + 2CO2(g)→ 2CO2(g) + 2CO(g) + O2(g) ∆H0= -788kJ/mol + 566kJ/mol
Subtracting the molecules that ar in both sides of the reaction:
2C(s) + O2(g) → 2CO(g) ∆H0 = -222kJ/mol
Write the relation of M3 with its multiples
Explanation:
HI friends good morning
A sample of gas has a volume of 20 cm³.The pressure is changed to 90 kPa at constant temperature,while the volume increases to 75 cm³.What was the original pressure of the gas?
Answer:
337.5kPa ~ 338kPa
Explanation:
Using the ideal gas law PV=nRT we have the following definitions from the problem:
V(initial) = 20cm³
P(initial) = ?kPa
V(final) = 75cm³
P(final) = 90kPa
Since we know that the number of moles of the sample did not change, nor did the temperature, nor does the ideal gas constant (R) we can rewrite this equation to state:
P(initial)V(initial) = nRT =P(final)V(final) ~ P(initial)V(initial) = P(final)V(final)
Rearranging this equation as we are solving for the initial pressure we find that:
P(initial) = (P(final)V(final))/V(initial)
P(initial) = ((90kPa)(75cm³))/20cm³
P(initial) = 337.5kPa ~ 338kPA