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Calculating Delta G at Nonstandard Conditions

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Language: en

00:00:03.569
all right we're today we're gonna do the
00:00:06.030 00:00:06.040 problem from chapter 17 where we're
00:00:07.950 00:00:07.960 calculating Delta G and for this
00:00:10.619 00:00:10.629 reaction at non-standard conditions at
00:00:12.600 00:00:12.610 25 degrees Celsius the non-standard
00:00:14.999 00:00:15.009 conditions are when the pressure of
00:00:16.850 00:00:16.860 ch3oh gas is 0.85 5 ATM pressure of
00:00:22.740 00:00:22.750 carbon monoxide gas is 0.125 atm and the
00:00:27.630 00:00:27.640 pressure pressure of hydrogen gas is
00:00:29.729 00:00:29.739 0.18 3 atm
00:00:32.360 00:00:32.370 to find Delta G at non-standard
00:00:35.190 00:00:35.200 conditions we use the equation Delta G
00:00:37.410 00:00:37.420 is equal to Delta G naught plus RT
00:00:39.960 00:00:39.970 natural log of Q where R is the gas
00:00:43.140 00:00:43.150 constant which is equal to 8.314 joules
00:00:46.740 00:00:46.750 per mole Kelvin and T is the temperature
00:00:49.530 00:00:49.540 in Kelvin so I had to convert 25 degrees
00:00:52.560 00:00:52.570 Celsius and add 273 to get 298 Kelvin to
00:00:57.270 00:00:57.280 find Delta G naught we use the equation
00:00:59.220 00:00:59.230 the sum of Delta G of the products minus
00:01:02.280 00:01:02.290 the sum of Delta G of the reactants for
00:01:05.280 00:01:05.290 this problem we use Delta G of carbon
00:01:08.520 00:01:08.530 monoxide gas plus 2 times Delta G of
00:01:11.910 00:01:11.920 hydrogen gas minus Delta G of ch3oh gas
00:01:17.510 00:01:17.520 always keep in mind you have to multiply
00:01:19.500 00:01:19.510 Delta G by the coefficient so you can
00:01:22.110 00:01:22.120 see that we multiplied hydrogen gas by 2
00:01:24.630 00:01:24.640 because it has a coefficient of 2 in the
00:01:26.940 00:01:26.950 reaction using the values from appendix
00:01:29.610 00:01:29.620 to be in the textbook we found our Delta
00:01:32.730 00:01:32.740 G naught to be equal 25 point 1
00:01:35.010 00:01:35.020 kilojoules per mole to calculate the
00:01:36.810 00:01:36.820 reaction quotient Q we use the law of
00:01:39.270 00:01:39.280 mass action which has the partial
00:01:41.010 00:01:41.020 pressure of the products over the
00:01:43.080 00:01:43.090 partial pressure of the reactants so we
00:01:46.200 00:01:46.210 had the partial pressure of carbon
00:01:47.700 00:01:47.710 monoxide gas times the partial pressure
00:01:50.340 00:01:50.350 pressure of hydrogen gas
00:01:52.310 00:01:52.320 you must square the value of hydrogen
00:01:54.810 00:01:54.820 gas because it has a coefficient of 2 in
00:01:57.270 00:01:57.280 the reaction we divide this by the
00:02:00.210 00:02:00.220 partial pressure of ch3oh gas and using
00:02:03.960 00:02:03.970 the given values from the problem we
00:02:05.999 00:02:06.009 found the reaction quotient to be four
00:02:07.890 00:02:07.900 point eight nine six times 10 to the
00:02:10.020 00:02:10.030 negative 3 to ensure we have similar
00:02:11.940 00:02:11.950 units you convert the gas constant 8.314
00:02:16.200 00:02:16.210 joules per mole Kelvin kilojoules which
00:02:18.810 00:02:18.820 gives us a final value of 8.314 times 10
00:02:22.500 00:02:22.510 the negative 3 kilojoules per mole
00:02:24.570 00:02:24.580 Kelvin so now we plug in our calculated
00:02:27.810 00:02:27.820 values into the original equation so we
00:02:30.630 00:02:30.640 have Delta G at non-standard conditions
00:02:32.370 00:02:32.380 is equal to 25 point 1 kilojoules plus
00:02:36.980 00:02:36.990 8.314 times 10 to the negative 3
00:02:39.540 00:02:39.550 kilojoules per mole Kelvin times 298
00:02:43.830 00:02:43.840 Kelvin times the natural log of 4 point
00:02:47.280 00:02:47.290 8 9 6 times 10 to the negative 3
00:02:50.660 00:02:50.670 calculating this we get Delta G is equal
00:02:53.070 00:02:53.080 to 25 point 1 kilojoules plus negative
00:02:56.280 00:02:56.290 13 point one seven nine kilojoules which
00:02:59.880 00:02:59.890 gives us our final answer which is Delta
00:03:02.430 00:03:02.440 G at non certain conditions is equal
00:03:04.050 00:03:04.060 eleven point nine kilojoules always make
00:03:06.630 00:03:06.640 sure you keep track of your significant
00:03:08.550 00:03:08.560 figures

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