Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25oC.?
Answer The change in the enthalpy of the system during a chemical reaction is equal to the change in the internal energy plus the change in the product of the pressure of the gas in the system and its volume. delta H = delta E + P* delta V [ at constant pressure ] delta V is change in volume that takes place during reaction. P* delta V is the work done on surrounding when the system expands by volume delt V at pressure P 2CO(g) + O2(g) → 2CO2(g) ∆H = -566.0kJ/mol 2 moles of CO2 at STP that is 273 k , 1 atm pr has 2*22.4 = 44.8 lits atm , the same at 298 deg k and constant pressure at 1 atm will have vol of 44.8*298/273 = 48.90 lits In this case the work done is - 1*48.9 lit-atm, the negative sign indicates there has been no expansion of volume during reaction since 3 moles of gases in reactants reduced to 2 moles in products So work done is - 48.9 lits-atm or - 48.9 * 0.101325 = - 4.95 kj [ 1 liter-atmosphere is equal to 0.101325 kilojoules (kJ)] . since, delta H = delta E + P* delta V [ at constant pressure ] -566 = delta E - 4.95 delta E = -561.05 kj
Calculate the entropy change when one mole of ice (H2O) melts at 0 degrees celcius and 1 atm.?
At the melting point, liquid is in equilibrium with the solid. That means that the free energy change for the reaction: H2O(s) -> H2O(L) is equal to zero. ΔG = 0 = ΔH - T*ΔS ΔS = ΔH/T For this reaction, ΔH is simply the enthalpy of fusion and T = 273K, so: ΔS_fusion = (6010J/mol)/(273K) = 22.02 J/(K*mol) If 1 mole of water melts, then the entropy change of the water is 22.02 J/K.
Calculate the enthalpy change...?
(25.0 g H2O) / (18.01532 g H2O/mol) = 1.3877 mol H2O (6.01 kJ/mol + 40.67 kJ/mol) x (1.3877 mol H2O) = 64.778 kJ to melt the ice and boil the water (2.09 J/g-K) x (25.0 g) x (0°C - (-4°C)) = 209 J to warm the ice to 0°C (4.18 J/g-K) x (25.0 g) x (100°C - 0°C) = 10450 J to warm the melted ice to 100°C (1.84 J/g-K) x (25.0 g) x (110°C - 100°C) = 460 J to heat the steam to 110°C 64778 J + 209 J + 10450 J + 460 J = 75897 J = 75.9 kJ total
How much amount of heat is required to convert 1 gram of ice at 0 degree Celsius into steam at 100 degree Celsius?
Ans) At 0℃ both ice and water exist. Here first ice will change to water but the temperature will be same i.e it will become water at 0℃. You may be wondering but this is due to latent heat which is hidden inside. We can find out heat required by this process :H = m×Lf (where m is mass in kg and Lf is latent heat of fusion and is value is fixes and is 3.35 × 10^5 J / kg for ice.)So the amount of heat required to change 1 gram of ice to water is :( 1g = 1/1000kg).001 × 3.35×10^5=335J. -(equation 1)Now after this on heating water will reach hoti 100 ℃. So about of heat required to change temperature is given by:H = mc∆T. (where m is mass in kg , c is specific heat and it's value is fixed . For water it is 4200J/kg ,for ice it is 2100 J/kg and for steam it is 2010J /kg. And ∆T is change in temperature required.)Here we have to change the temperature of water ,so we will apply for water.Heat required:H= .001× 4200× (100–0)=420J (equation 2)Now on heating further water will change to steam . And the formula to find is mLv where m is mass in kg , Lv is latent heat of vaporization and it's value is 22.5J/kg for water.)H= .001×22.5×10^5.=2250J. (equation 3)On adding equation (1),(2),(3) we will get the total heat required.335J + 420J + 2250 J = 3005 J.You can change this to calorie as1 calorie = 4.184 J.So 3005 J= 3005 ×4.184 calorie =718.21 calorie.
Why is the change in entropy of the vaporization of water 0 at 373 K?
The change in entropy of water cannot be 0 during vaporization. By definition, entropy is a measure of the "disorder" of a system or randomness.During vaporization the randomness of the system increases, so change in entropy will always have a positive value and cannot be 0.Change in Entropy at Vaporization is given by, [math]ΔHvap / T[/math]For example, when 1 g of liquid water evaporates at 373K(ΔHvap=40.63kJ/mol)[math]ΔSvap=ΔHvap / T[/math][math]ΔSvap[/math]=40.63×1000(J/mol) / 373 K[math]ΔSvap=[/math][math]109J/K−mol[/math]The above value is the change in entropy for 1 mol of water i.e. 18 grams of water. But we need the value for 1 gram of water.So, Entropy change for evaporation of 1 g of water =( 109×1) / 18=6.05 J/K
How much energy is needed to convert 20 grams of ice at -10 degree C into steam at 110 degree C?
Temp. change = Mass(g) x SH (J/g/°C) x °C ΔT. Phase change = Mass x LH. You have the SH of steam as 4.2J/g/°C...it's 2.01J/g/°C. *..Ice at -10°C to 0°C = 20 x 2.1 x 10 = 0.420kJ. *..Ice at 0°C to water at 0°C = 20 x 336 = 6.7kJ. *..Water at 0°C to 100°C = 20 x 4.2 x 100°C = 8.4kJ *..Water at 100°C to steam at 100°C = 20 x 2.3 = 46kJ *..Steam at 100°C to 110°C = 20 x 2.01 x 10 = 0.402kJ Total heat required (kJ) = 0.42 + 6.7 + 8.40 + 46. + 0.402 = 61.92kJ.
How much heat is involved in converting 1 mol of steam at 160 degrees Celsius to ice at -45 degrees Celsius?
Steam q1 = mc*delta t t1 = 100oC t2 = 160oC c = 2.01 J/oC*mole m = 1 mole = 18 grams (2H + 0 = 2 + 16 = 18) q1 = 18 * 2.01*(160 - 100) q1 = 2170 J Stream to water =========== q2 = n*H n = 1 mole H = 40680 J/mol When you look this up, it is the heat of vaporization. q2 = 40580 J/mol * 1 mol = 40680 Water from 100 to zero ================= 1 mol = 18 grams c = 4.19 J/g oC delta t = 100 - 0 = 100oC q3 = 18*4.19*100 q3 = 7542 J Water to Ice n = 1 mol Heat of Fusion for ice = 6000 J /mol q4 = n*Hf q4 = 1*6000 J/mole q4 = 6000 J Ice from 0 to -45oC q5 = m*c*(0- - 45) q5 = 18 grams * 4.19 J/oC * g * 45 q5 = 3394. The total heat lost = q1 + q2 + q3 + q4 + q5 = 59785 J Notice that for phase changes (steam to water and then water to ice) you use moles. For heat transfer but no phase change, you change the moles to grams.
Change of state - Ice to Steam?
1 mole of ice (0 c)to 1 mole of water (0 C): Enthalpy change of fusion= 6.01 KJ/mol seeing as the standard unit is in moles u can easily convert by simply dividing by 18 (seeing as 1 mole of water is approximately 18 grams): answer: 0.334 KJ or 334J or 80 calories 1 kg of water (0 C) to 1 kg of water (100 C) : Specific heat capacity= 4.1855 KJ/kg/K multiply this number by 100 (for every degree increase) and divide it by a 1000 since its unit is based on a kilogram. answer: 0.41855 KJ or 418.55J or 100 calories 1 mole of water (100 C) to 1 mole of steam (100 C): enthalpy change of vaporisation= 40.68 KJ/mol again divide by 18. Answer: 2.26 KJ or 2260J or 540 calories (all numbers are approximations and measurements standard temperature and pressure)