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What is the value for G at 100 K if H equals 27 kJ mol and S equals 0.09 kJ molK?
To find the value of G at 100 K, you can use the equation ΔG = ΔH - TΔS. Plugging in the values, you get ΔG = 27 kJ/mol - (100 K)(0.09 kJ/molK) = 18 kJ/mol. Therefore, the value for G at 100 K would be 18 kJ/mol.
What is the value for G at 100 K if H 27 kJmol and S 0.09 kJ(molK)?
The equation relating G, H, and S is G = H - TS, where T is the temperature in Kelvin. Plugging in the values given, G = 27 kJ/mol - 100 K * 0.09 kJ/(molK) = 27 kJ/mol - 9 kJ/mol = 18 kJ/mol. So, the value for G at 100 K is 18 kJ/mol.
What is the value for g at 5000 k if h -220 kj mol and s equals -0.05 kj molk?
-18 kj/mol
What is the internal energy of 4.50 mol of N2 gas at 253C To solve this problem use the equation Remember that R 8.31 J(molK) and K C plus 273.?
The internal energy of a gas is given by the equation U = nCvT, where n is the number of moles, Cv is the molar heat capacity at constant volume, and T is the temperature in Kelvin. For N2 gas, Cv is 20.8 J/(molK). Plugging in the values, U = (4.50 mol) * (20.8 J/(molK)) * (253+273) K. Solving gives you the internal energy.
Which statement describes a reaction at 298 K if H 31 kJ mol S 0.093 kJ molK?
It is not spontaneous.
What is the value for G at 100 K if H equals 27 kJ mol and S equals 0.09 kJ molK?
To find the value of G at 100 K, you can use the equation ΔG = ΔH - TΔS. Plugging in the values, you get ΔG = 27 kJ/mol - (100 K)(0.09 kJ/molK) = 18 kJ/mol. Therefore, the value for G at 100 K would be 18 kJ/mol.
What is the value for G at 100 K if H 27 kJmol and S 0.09 kJ(molK)?
The equation relating G, H, and S is G = H - TS, where T is the temperature in Kelvin. Plugging in the values given, G = 27 kJ/mol - 100 K * 0.09 kJ/(molK) = 27 kJ/mol - 9 kJ/mol = 18 kJ/mol. So, the value for G at 100 K is 18 kJ/mol.
What is the value for g at 5000 k if h -220 kj mol and s equals -0.05 kj molk?
-18 kj/mol
What can be said about a reaction with H equals -890 kJ mol and S equals -0.24 kJ molK?
It is spontaneous at 2000 K.
What is the value of G at 500 K if H27kJMol and S0.09kJ(molK)?
To find the Gibbs free energy change (ΔG) at 500 K, we can use the equation ΔG = ΔH - TΔS. Given that ΔH = -27 kJ/mol and ΔS = 0.09 kJ/(mol·K), we first convert the temperature to Kelvin (which is already given as 500 K). Then, substituting the values: ΔG = -27 kJ/mol - (500 K × 0.09 kJ/(mol·K)) = -27 kJ/mol - 45 kJ/mol = -72 kJ/mol. Thus, the value of G at 500 K is -72 kJ/mol.
If a reaction has an enthalpy of -54.32 kJ/mol and an entropy of -354.2 J/(K*mol), what is the Gibbs free Energy at 54.3(degrees c)?
DeltaG = DeltaH - TDeltaS dG = -54.32 kJ/mol - (54'32+273)K(-354.2J/molK) NB Thevtemperature is quoted in Kelvin(K) and the Entropy must be converted to kJ by dividing by '1000'/ Hence dG = - 54.32kJ/mol - (327.32K)(-0.3542 kJ/molK) NB The 'K' cancels out. Then maker the multiplication dG = -54/32 kJ/mol - - 115.94 kJ/mol Note the double minus; it becomes plus(+). Hence dG = -54.32kj/mol + 115.94 kJ/mol dG = (+)61.61 kJ/mol Since dG is positive, the reaction is NOT thermodynamically feasible.
What is the internal energy of 4.50 mol of N2 gas at 253C To solve this problem use the equation Remember that R 8.31 J(molK) and K C plus 273.?
The internal energy of a gas is given by the equation U = nCvT, where n is the number of moles, Cv is the molar heat capacity at constant volume, and T is the temperature in Kelvin. For N2 gas, Cv is 20.8 J/(molK). Plugging in the values, U = (4.50 mol) * (20.8 J/(molK)) * (253+273) K. Solving gives you the internal energy.
What volume will 1.44 kg of oxygen gas occupy at the planet's aatmospheric pressure?
PV=nRT P=Pressure V=Volume n=mols R=Ideal gas constant (8.3145 m3·Pa/(mol·K)) T=Temperature Atmospheric pressure is highly variable, based upon weather and altitude. Temperature fluctuates as well. Assuming: Pressure 100 KPa Temperature 25 C 100 KPa x V = (1.44kg/32g/mol)(8.3145m3Pa/(molK))(25+273) 100 KPa x V = 45 mol x 8.3145 m3Pa/(molK) x (298) V = 45 x 8.3145 x 298 / 100000 V = 1.115 m3
Use the reaction I2(s) I2(g), H = 62.4 kJ/mol, S = 0.145 kJ/(molK), for this question What can be said about the reaction at 500 K?
It is spontaneous.
Which statement describes a reaction at 298 K if H 31 kJ mol S 0.093 kJ molK?
It is not spontaneous.
How may Liters in a Volume of 1.20 mol of gas at 61.3kPa and 25 degrees C?
The volume can be calculated using the ideal gas law equation, V = (nRT) / P, where n is the number of moles (1.20 mol), R is the gas constant (8.31 LkPa/molK), T is the temperature in Kelvin (25 + 273 = 298 K), and P is the pressure (61.3 kPa). Substituting these values into the equation gives V = (1.20 mol * 8.31 LkPa/molK * 298 K) / 61.3 kPa ≈ 49.1 L.
What is the mole fraction of 18 mol NaCl in a solution of 100 mol water?
The mole fraction of NaCl in the solution is calculated by dividing the moles of NaCl by the total moles of NaCl and water. In this case, the mole fraction of 18 mol NaCl in a solution of 100 mol water is 0.15 (18 mol NaCl / (18 mol NaCl + 100 mol water)).
