Answer:
ΔH = 53.28 kJ
Explanation:
Solving this kind of problems is quite straight forward. What is need is to manipulate the reactions by multiplying the coefficients and reversing if necessary the reactions so that when we add the reactions together at the end we will arrive to the desired equation we need to obtain the enthalpy.
The reaction we need to calculate the enthalpy is
3C(s) + 3H2(g) → C3H6(g) ΔH = ?
If we take the 2nd reaction and multiply it by three, the inverse of first one multiplied by 1/2 and three times the 3rd we will be get the desired equation and its enthalpy:
3 C(s) + 3 O2(g) → 3 CO₂(g) ΔH= 3 x ( -393.51 kJ)
3 CO₂(g) + 3 H₂O(l) → C₃H₆(g) + 9/2 O₂(g) ΔH= 1/2 x ( 4182.6 kJ)
3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l) ΔH= 3 x ( -285.83 kJ)
3C(s) + 3H2(g) → C3H6(g)
Notice how the mole O2 cancel because they are in diferent sides of the equation. Also we changed the changed of the second since we inverted it.
ΔH = 3 x ( -393.51 kJ) + 1/2 x ( 4182.6 kJ) + 3 x ( -285.83 kJ)
ΔH = - 1180.53 + 2091.3 - 857.49 = 53.28 kJ
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m = 0.17 kg
v=37m/s
h=6.62*10^-34m²kg/s
x= h/mv
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