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kaheart [24]
3 years ago
7

Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.960 atm when 24.0 g of glucose is consumed in this re

action.
Chemistry
1 answer:
weqwewe [10]3 years ago
6 0
The balanced equation for the above reaction is as follows;
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O
molar ratio of glucose to CO₂ is 1:6
the mass of glucose reacted - 24.0 g
Number of glucose moles reacted - 24.0 g /180 g/mol = 0.133 mol
Number of CO₂ formed when 0.133 mol of glucose reacted -0.133 x 6 = 0.798 mol
since CO₂ is a gas we can use the ideal gas law equation to find the volume occupied 
PV = nRT
where P - pressure - 0.960 atm x 101 325 Pa / atm = 97 272 Pa
n - number of moles - 0.798 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 273 + 37 °C = 310 K
substituting the values in the equation
97 272 Pa x V= 0.798 mol x 8.314 Jmol⁻¹K⁻¹ x 310 K
V = 21.1 dm³
volume occupied by CO₂ is 21.1 dm³
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The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

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We get,

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