Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.
1) <span>A decrease in the concentration of the reactants:
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.
2) </span><span>A decrease in the surface area of the products:
Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.
3) </span><span>An increase in the temperature of the system:
An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.
4) </span><span>An increase in the activation energy of the forward reaction:
An increase in Activation energy will decrease the rate of reaction, either it is forward or reverse. So this is incorrect.
Result:
Hence, the correct answer is,"</span>An increase in the temperature of the system".
In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.
O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13
CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044
Answers: O2 = 0.13
CO2 = 0.044
