CH3OH(g) CO(g) + 2 H2(g) H = +90.7 kJ calculate E when 830.0 g of CH3OH(g) compl

Question

CH3OH(g) CO(g) + 2 H2(g) H = +90.7 kJ calculate E when 830.0 g of CH3OH(g) completely reacts at a constant temperature of 300 K and constant pressure of 0.95 atm. R = 8.314 J/mol*K and R = 0.08206 atm*L/mol*K

Explanation / Answer

1) V1/T1=V2/T2 --> V2 = V1x(T2/T1) = 300cm³ x (333K/303K) = 330cm³ (assuming constant P) 2) V1/T1=V2/T2 -->V1=V2x(T1/T2) = 1.0Lx(300K/100K) = 3.0L 3) you have 2x as many moles of antacid ---> 2x as many moles of gas...V1/n1 = V2/n2 ---> V2 = V1 x (n2/n1).. double the moles, double the volume... so gas from 4 tablets > gas from 2 tablets. 4) P1/T1=P2/T2 ---> P2=P1x(T2/T1)..if T was left in °C, P2=100kPax(10°C/60°C) = 17kPa. T MUST be in absolute temperature. either K or R. never °F nor °C. units of P do not matter.

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