What is the value of G (in kJ/mol) at 100 C for the following reaction: Ca(s) +

Question

What is the value of G (in kJ/mol) at 100 C for the following reaction: Ca(s) + Cl2(g) = CaCl2(s). At 25 C, H = -795.8 kJ/mol; S = -159.8 J/K-mol and G = -748.1 kJ/mol. Assume H and S do not vary with temperature. Answer in kJ/mol. -736.1, 5.88E4, -855.4, -779.8, 1.52E4

Calculate the change in entropy as the 2.00  lbs of liquid butane (CH3CH2CH2CH3) inside a camping stove fuel tank is turned into gaseous butane at a constant temperature of 5.00 C on a brisk morning in the wilderness. For butane, the delta-H of fusion is 8.00 kJ/mol and the delta-H of vaporization is 24.0 kJ/mol. The units are J/K.

Given the following values of H, S and T, calculate G (in kJ). The values are: H = -370 kJ, S = 230 J/K, and temperature = 55.0 C.

Thank you

Explanation / Answer

we have to use G = H - TS

G 100 ºC = H - TS

-795,800 J/mole - (373 K)(-159.8 J/K mole) =

= -736194.6 J/mole

= -736.194 kj/mol

sorry for second one i am not very confident

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