203a4a3cdfb5","label":null,"content":"Use the graph to calculate a value for the activation energy ($E_a$), in $\\text{kJ mol}^{-1}$, for this reaction.\n\nThe gas constant $R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$.","markscheme":"- Identifies two sets of coordinates from the straight line graph, for example $(0.00120, -1.0)$ and $(0.00180, -9.4)$ M1\n- Calculates the gradient of the line: $m = \\frac{-9.4 - (-1.0)}{0.00180 - 0.00120} = -14000\\text{ K}$ (allow gradient range $-13500\\text{ to } -14500\\text{ K}$) M1\n- Recognises that the gradient $m = -\\frac{E_a}{R}$ or uses $E_a = -\\text{gradient} \\times R$ M1\n- Calculates the value of $E_a$ in J: $E_a = 14000 \\times 8.31 = 116340\\text{ J mol}^{-1}$ A1\n- Divides by $1000$ to convert to $\\text{kJ mol}^{-1}$: $116\\text{ kJ mol}^{-1}$ (accept any answer in the range $112\\text{ to } 120\\text{ kJ mol}^{-1}$ based on reading tolerance) A1","marks":5,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"gradient_method","label":"Gradient Method","steps":[{"type":"text","order":0,"title":"Relate gradient to Arrhenius equation","content":"The Arrhenius equation is given as:\n\n$$\\ln k = \\frac{-E_a}{RT} + \\ln A$$\n\nBy rearranging this into the form of a straight line, $y = mx + c$, where $y = \\ln k$ and $x = \\frac{1}{T}$:\n\n$$\\ln k = \\left(-\\frac{E_a}{R}\\right)\\left(\\frac{1}{T}\\right) + \\ln A$$\n\nFrom this, we can see that the gradient ($m$) of the straight line is equal to:\n\n$$m = -\\frac{E_a}{R}$$","highlights":[{"text":"\\ln k = \\frac{-E_a}{RT} + \\ln A","location":"specification"}]},{"type":"text","order":1,"title":"Select coordinates from graph","content":"To calculate the gradient of the line of best fit, select two widely spaced points directly on the line of best fit.\n\nFrom the given graph, we can choose two clear points:\n* Point 1: $(x_1, y_1) = (0.00120, -1.0)$\n* Point 2: $(x_2, y_2) = (0.00180, -9.4)$"},{"type":"text","order":2,"title":"Calculate the gradient","content":"Now, compute the gradient ($m$) using the gradient formula:\n\n$$m = \\frac{\\Delta y}{\\Delta x} = \\frac{y_2 - y_1}{x_2 - x_1}$$\n\nSubstituting our chosen coordinates:\n\n$$m = \\frac{-9.4 - (-1.0)}{0.00180 - 0.00120} = \\frac{-8.4}{0.00060} = -14000\\text{ K}$$\n\n*(Note: An acceptable range for the gradient depending on your precise reading of the graph is $-13500\\text{ to } -14500\\text{ K}$)*","calculator":[{"gdc":{"expression":"(-9.4 - (-1)) / (0.0018 - 0.0012)","instructions":"Enter the fraction (-9.4 - (-1)) / (0.0018 - 0.0012) into your scientific or graphing calculator."},"ti84":{"expression":"(-9.4 - (-1)) / (0.0018 - 0.0012)","instructions":"Type (-9.4 - (-1)) / (0.0018 - 0.0012) and press ENTER."},"desmos":{"expression":"(-9.4 - (-1)) / (0.0018 - 0.0012)","instructions":"Type (-9.4 - (-1)) / (0.0018 - 0.0012) into the input line."},"description":"Calculate the gradient using standard division."}]},{"type":"text","order":3,"title":"Calculate activation energy in Joules","content":"Since $m = -\\frac{E_a}{R}$, we can rearrange the equation to solve for the activation energy ($E_a$):\n\n$$E_a = -m \\times R$$\n\nSubstitute the gradient value ($m = -14000\\text{ K}$) and the gas constant ($R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}$):\n\n$$E_a = -(-14000\\text{ K}) \\times 8.31\\text{ J K}^{-1}\\text{ mol}^{-1} = 116340\\text{ J mol}^{-1}$$","calculator":[{"gdc":{"expression":"14000 * 8.31","instructions":"Multiply 14000 by 8.31."},"ti84":{"expression":"14000 * 8.31","instructions":"Type 14000 * 8.31 and press ENTER."},"desmos":{"expression":"14000 * 8.31","instructions":"Enter 14000 * 8.31."},"description":"Multiply the positive gradient value by the gas constant."}],"highlights":[{"text":"R = 8.31\\text{ J K}^{-1}\\text{ mol}^{-1}","location":"specification"}]},{"type":"text","order":4,"title":"Convert answer to kJ/mol","content":"The question asks for the activation energy in $\\text{kJ mol}^{-1}$. Convert from $\\text{J mol}^{-1}$ to $\\text{kJ mol}^{-1}$ by dividi

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