In the adiabatic, reversible expansion of 0.015 mol Ar, from 1.5 L to 2.0 L. Initial temperature of Ar is 30 deg C. Calculate the work done of in this expansion. [Given : Molar heat capacity of Ar at constant volume is 12.48 J / (K mol) , (Cv,m/R) = 1.501
We know, c = Cvm/R = 1.501 Cvm = 12.48 J / (K mol) Initial T = 30 deg C Moles of Ar = 0.015 mol Initial volume (Vi) = 1.5 L Final volume (Vf) = 2.0 L We need, Conversion of initial temperature (Ti) in K Final Temperature (Tf) Delta T ( Tfinal – T initial ) Conversion of T into K. T in K = 30.0 + 273.15 = 303.15 K Calculation of final T Tf = Ti x ( Vi/Vf)^(1/c) Tf = (303.15 K ) x ( 1.5 L/2.0 L )^(1/1.501) Tf=250.28 K Calculation of delta T Delta T = Tf-Ti = 250.28K – 303.15 K =-52.9 K Work done W = n x Cvm x Delta T = 0.015 mol x 12.48 J / (K mol ) x (-52.9 K ) = -9.90 J Work done in this expansion is -9.90 J (Negative sign shows that work is done by system)
Why the proton in acetylene is more acidic than ethylene?
Acidic nature of proton depends on the strength of acid. The pron in acetylene is bonded to sp hybridized carbon, where as in ethylene its bonded to the sp2 hybridized carbon. Percent of S in sp hybridization is more than sp2. S orbital is close to nucleus so electrons are more attracted to the nucleus which results into more electronegativity of carbon in sp than carbon in sp2. More electronegative atom in the bond has tendency to pull the electron density towards itself and thus in C-H bond H becomes electron deficient. In acetylene, such a more electron positive H can liberated easily than sp2. We know the substance that liberate protons is called as acid and the strength of liberation is more in acetylene than ethylene therefore proton in acetylene is more acidic than ethylene.
Water and carbon dioxide are the products in the combustion reaction of methane. If 30.1 g of methane gas is reacted with 45.0 g O2, which is the limiting reactant? How many grams of water will be formed?
We know, The chemical reaction > 30.1 g CH4(g) > 45.0 g O2(g) We need, > Balanced chemical equation > Moles of CH4(g) > moles of O2(g) Balanced equation is, CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Calculation of moles In order to calculate moles, we need to obtain the molar masses. CH4 16.0426 g/mol O2 15.9980 g/mol 30.1 g CH4 x (1 mol CH4 / 16.0426 g CH4) = 1.876 mol CH4 45.0 g O2 x (1 mol O2 / 15.998 g O2) =2.813 mol O2 Limiting reactant is the one which is consumed first. We compare the mole ratio. 1 mol CH4 : 2 mol O2 ( 1 mol CH4/ 2 mol O2) Moles of O2 required to react with 1.876 mol CH4 = 1.876 mol CH4 x 2 mol O2 / 1 mol CH4 = 3.75 mol O2 Thus 3.75 moles of O2 are required to react completely with 1.876 mol CH4. Since only 2.813 moles of O2 are present actually, the amount of O2 is limiting. Calculation of mass of H2O formed. Because limiting reactant is O2, we must use the amount of it to calculate the mass of H2O. Mol ratio between O2 and H2O 2 mol O2 / 2 mol H2O Moles of H2O formed. n H2O = 2.813 mol O2 x ( 2 mol H2O / 2 mol O2) = 2.313 mol H2O Using molar mass of H2O, we can calculate the mass of H2O H2O 18.0148 g/mol m H2O = 2.813 mol H2O x (18.0148 g H2O / 1 mol H2O) = 50.7 g H2O Mass of water formed is 50.7 g