Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
The amount of heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 17.0 L of H2 at 75°C and 744 torr is;
Q = -37.58 KJ
We are given the reaction as;
CO(g) + 2H2(g) = CH3OH(l)
We are also given;
Volume of CO = 15 L
Temperature of CO = 85°C = 355 K
Pressure of CO = 112 KPa = 1.10535 atm
Volume of H2 = 17 L
Temperature of H2 = 75°C = 345 K
Pressure of H2 = 744 torr = 0.97895 atm
Number of moles of CO is gotten from the formula;
n = PV/RT
n = (1.10535 × 15)/(0.0821 × 355)
n = 0.5689 mol
Number of moles of H2 is;
n = (0.97895 × 17)/(0.0821 × 345)
n = 0.58755 mol
From tables, the enthalpy of formation and reaction of CO, H2 and CH3OH are;
ΔHf° CO = -110.5 kJ/mol
ΔHf° H2 = 0 kJ/mol
ΔHf° CH3OH = -238.42 KJ/mol
From Hess's law, we know that;
ΔH_rxn = Σ(n × ΔHf°)_products - Σ(n × ΔHf°)_reactants
Now from the balanced reaction, we see that 1 mole of CO reacts with 2 mole of H2 to give 1 mol of CH3OH.
Thus;
ΔH_rxn = (1 × -238.42) - (1 × -110.5)
ΔH_rxn = −127.92 KJ/mol
Formula for quantity of heat is;
Q = n × ΔH_rxn
Putting 0.58755/2 mol because from the balanced equation given, 2 moles of H2 are used for every mile of CO and thus it means H2 is limiting..
Thus;
Q = 0.58755/2 × -127.92
Q = -37.58 KJ
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