The universal gas constant, R = 8.314 J/mol. K. In other unit system, R = 0.08214 L.atm/ mol. K. Using unit conversion, show how the first value for R can be converted to the second value.

Determine the equilibrium constant for a reaction at 200 K if ∆G° =23.5 kJ/mol. (R = 8.314 J/mol ･ K) Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

Science; Chemistry; Chemistry questions and answers; a. Determine K for a reaction at 200 K if ∆G° =21.1 kJ/mol. (R = 8.314 J/mol ･ K) b.

Convert the universal gas constant R = 8.314 J/mol/K into units of calories/mol/R. Note that the unit R = Rankine, which is absolute temperature on the Fahrenheit scale, where absolute zero is -459.7ºF. So, to convert from ºF to R, add 459.7, so 1 K = 1.8 R.

What is K for a reaction if ∆G° =-287.0 kJ/mol at 25 °C (or 298 K)? (R = 8.314 J/mol ･ K) At 355 K, ∆G = -5.7 kJ/mol for the reaction A (g) → 2 B (g). If the partial pressures of A and B are 3.67 atm and 0.110 atm respectively, what is the standard free energy for …

Question: Show for the ideal gas constant R that the following units equivalency is true, R 8.314 J/K-mol and R 8.314 kPa-l/K-mol (lliters, K-absolute temperature, J-Joules, Pa Pascal) Show transcribed image text

Answer to R = 8.314 J-mol-K-1 = 8.314 mPa.mol-1 K-1 83.14 cm. R = 8.314 J-mol-K-1 = 8.314 mPa.mol-1 K-1 83.14 cm bar mol-.K-1 = 8314 cm. kPa.mol-1.K-1 82.06 cm (atm).mol-'K-' = 62, 356 cm (torr) mol K-1 1.987 (cal)-mol K- = 1.986 (Btu)(lb mole)'(R)-1 0.7302 (ft)' (atm) (lb mol)--(R)-? = 10.73 (ft)' (psia) (lb mol)--(R)-1 1545 (ft) (lbf) (lb mol)- …

(R = 8.314 J/mol ･ K) There are 2 steps to solve this one. Solution. Step 1. Here's the solution: Based ...

For the Ideal gas calculation, use R = 8.314 J / mol. (or similar variation); and molar mass of H 2 O = 18.02 g / mol. Consider steam at 20 0 ∘ C and 5 bar. (Give all answers to the nearest thousandth.) The specific volume based on the steam tables is m 3 / k g. The specific volume based on the ideal gas law is m 3 / kg

The boiling point of butane at 1 atm pressure is -1.0 °C and its ∆H(vap) is 22.44 kJ/mol. R = 8.314 ×10⁻³ kJ/mol･K. Calculate the pressure (in atm) of the butane in the lighter at 19.5 °C. Liquid butane is used in cigarette lighters.