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To calculate the Gibbs free energy (G) at 700 K, we can use the formula: ( G = H - TS ). Given that ( H = -92 , \text{kJ/mol} ) and ( S = -0.199 , \text{kJ/(mol·K)} ), we first calculate ( TS = 700 , \text{K} \times -0.199 , \text{kJ/(mol·K)} = -139.3 , \text{kJ/mol} ). Then, substituting into the Gibbs equation:
[ G = -92 , \text{kJ/mol} - (-139.3 , \text{kJ/mol}) = 47.3 , \text{kJ/mol}. ]
Thus, the value for G at 700 K is 47.3 kJ/mol.
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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.
Determine the empirical formula what compound if a sample contains 0.104 mol K 0.052 mol C and 0.156 mol O?
To find the empirical formula, we need to determine the ratio of each element in the compound. First, find the moles of each element: K = 0.104 mol C = 0.052 mol O = 0.156 mol Next, divide each mole value by the smallest mole value to get the ratio: K = 0.104 mol / 0.052 mol = 2 C = 0.052 mol / 0.052 mol = 1 O = 0.156 mol / 0.052 mol = 3 Therefore, the empirical formula is K2CO3.
At which temperature would a reaction with h -92 kj mol s-199 kj molk be spontaneous?
To determine the temperature at which the reaction becomes spontaneous, we can use the Gibbs free energy equation: ΔG = ΔH - TΔS. For a reaction to be spontaneous, ΔG must be less than 0. Given ΔH = -92 kJ/mol and ΔS = -199 kJ/(mol·K), we can set up the equation -92 kJ/mol - T(-199 kJ/(mol·K)) < 0. Solving for T gives T > 0.462 K, indicating that the reaction will be spontaneous at temperatures above this value.
What is the accepted value for the heat of solution NH4Cl?
The accepted value for the heat of solution of NH4Cl (ammonium chloride) is approximately +14.8 kJ/mol. This value represents the energy released or absorbed when one mole of NH4Cl dissolves in water at standard conditions.
If 5 moles of hydrogen gas is heated from 30 degrees C to 60 degrees C at constant pressure how much heat was added to the system?
The heat added can be calculated using the formula: q = nCpΔT, where q is the heat added, n is the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is the temperature change. Given n=5 moles, Cp for hydrogen gas is approximately 28.8 J/molK, and ΔT = 60°C - 30°C = 30 K, the heat added would be q = 5 mol * 28.8 J/molK * 30 K = 4320 J.
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 100 K if H 27 kJ mol and S 0.09 kJ molK?
G=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.
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 internal energy of 4.00 mol of diatomic nitrogen gas n2 at 455k?
The internal energy of a gas is given by the equation: U = (3/2) * n * R * T, where n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. For diatomic nitrogen gas (N2), the gas constant is 8.314 J/(molK). Plugging in the values: U = (3/2) * 4.00 mol * 8.314 J/(molK) * 455 K = 55957.76 J.
